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1.

`(2n+7) lt (n+3)^(2)`

Answer» For n=1
L.H.S `=2xx1+7=9`
R.H.S. `=(1+3)^(2)=16`
`:. " "L.H.S. lt R.H.S.`
`rArr` Given statement is true for n=1
Let given statement be true for n=k
`:. " "2k+7 lt (k+3)^(2)`
for n=K+1
`2(k+1)+7 =(2k+7)+2`
`lt (k+3)+2`
[From inequation (1)]
`=K^(2)+6k+11`
`lt (k^(2)+6K+11)+(2k+5)`
`lt K^(2)+8K+16lt (k+4)^(2)`
` rArr 2(k+1) +7 lt (K+4)^(2)`
`rArr` Given statement is also true for n=K+1
Hence from the principle of mathematical induction P (n) is true for all natural numbes n.
Discussion
2.

If `P(n):2nltn!,ninN` then P(n) is true for all `n≥` . . . . . . . . . .

Answer» Given that `P(n):2nltn!,ninN`
For n=1, `2lt!` [false]
For n=2, `2xx2lt!4lt2` [false]
For n=3, `2xx3lt!`
`6lt3!`
`6lt3xx2xx1`
`(6lt6)` [false]
For n=4, `2xx4lt4!`
`8lt4xx3xx2xx1`
`(8lt24)` [true]
`For n=5, 2xx5lt5!`
`10lt5xx4xx3xx2xx1`
`(10lt120)` [true]
Hence, P(n) is for all `nle4`. [true]
Discussion
3.

Using principle of mathematical induction, prove that `7^(4^(n)) -1` is divisible by `2^(2n+3)` for any natural number n.

Answer» Let `P(n) = 7^(4^(th)) - 1` be divisible by `2^(2n+3)`
`P(1) = 7^(4) - 1 = (7^(2) - 1) (7^(2) + 1)`
` = 48 xx 50 = 32 xx 75`
`= 2^(5) xx 75`.
which is divisible by `2^(2xx1+3)`
Let us assume that the result is true for n` = k`.
i.e, `7^(4^(k)) - I` is divisible by `2^(2k+3)` but not by `2^(2k+4)`.
`rArr 7^(4^(k)) - 1 = 2^(2k+3)`m , where m is some odd natural number
Now, `7^(4^(k+1)) - 1 = (7^(4^(k)))^(4-1)`
`= ((2^(2k+3)m+1)^(2) +1)((2^(2k+3)m+1)^(2)-1)`
`= ((2^(2k+3)m+1)^(2) +1)(2^(2k+3)+2)(2^(2k+3)m)`
`= (2^(4k+6)m^(2) + 2^(2k+4) m + 2)(2^(2k+3)m + 2) (2^(2k+3)m)` ltbr `= 2^(2k-5)(2^(4k+5)m^(2) + 2^(2k+3) m + 1)(2^(2^(2k+2)m +1) (m),`
Which is divisivble by `2^(2k+5)`
Thus, `P(k + 1)` is true whenever `P(k)` is true.
So, by the principle of mathematical induction, `P(n)` is true for any natural number n.
Discussion
4.

For every positive integer n, prove that `7^n-3^n`is divisible by 4.

Answer» `7^n - 3^n = 4t`
a) `n=1`
b) `n=k, n=k+1`
a) n=1, LHS`7^n - 3^n= 7-3 = 4`
RHS`= 4`
b) Assumption `n=k`
`7^n - 3^n = 4k`
`n= k+1`
LHS: `7^(k+1) - 3^(k+1)`
`= 7*7^k - 3*3^k`
`= 4*7^k + 3 * 7^k - 3*3^k`
`= 4(7^k) + 4t`
`= 4(7^k + t)`
Answer
Discussion
5.

What is the total number of proper subsets of a set containing n elements?

Answer» Let P(n) : number of subset of a set containing n distinct elements `2^(n)`, for all n `in N`.
Step I We observe that of P(1) is true, for n=1.
Number of subsets of a set contain 1 element is `2^(1)=2`, which is true.
Step II Assume that P(n) is true for n=k.
P(k) : Number of subsets of a set containing k distinct elements is `2^(k)`, which is true.
Step III To prove P(k+1) is true, we have to show that
P(k+1): Number of subsets of a set containing (k+1) distinct elements is `2^(k+1)`
We know that, with addition of one element in the set number of subsets become double.
`:.` Number of subsets of a set containing (k+1) distinct elements `=2xx2^(k)=2^(k+1)`.
So, P(k+1) is true. Hence, P(n) is true.
Discussion
6.

Prove that `n^5/5 + n^3/3+(7n)/(15)` is a natural number.

Answer» Consider the given statement
`P(n):(n^(5))/(5)+(n^(3))/(3)+(7n)/(15)` is a natural number, for all n `inN`.
Step I We observe that P(1) is true.
`P(1):((1)^5)/(5)+(1)^(3)/(3)+(7(1))/(15)=(3+5+7)/(15)=(15)/(15)=1`, which is a natural number. Hence, P(1) is true.
Step II Assume that P(n) is true, for n=k.
`P(1):((k)^5)/(5)+(k)^(3)/(3)+(7k)/(15)` is natural number. Step III Now, to prove P(k+1) is true.
`(k+1)^(5)/(5)+(k+1)^(3)/(3)+(7(k+1))/(15)`
`=(k^(5)+5k^(4)+10k^(2)+5k+1)/(5)+(k^(3)+1+3k(k+1))/(3)+(7k+7)/(15)`
`=(k^(5)+5k^(4)+10k^(2)+5k+1)/(5)+(k^(3)+1+3k^(2)+3k)/(3)+(7k+7)/(15)`

`=(k^(5))/(5)+(k^(3))/(3)+(7k)/(15)+(5k^(4)+10k^(3)+10k^(2)+5k+1)/(5)+(3k^(2)+3k+1)/(3)+(7k+7)/(15)`
`=(k^(5))/(5)+(k^(3))/(3)+(7k)/(15)+K^(4)+2k^(3)+2k^(3)+2k^(2)+k+k^(2)+(1)/(5)+(1)/(3)+(7)/(15)`
`=(k^(5))/(5)+(k^(3))/(3)+(7k)/(15)+K^(4)+2k^(3)+3k^(3)+2k+1`, which is a natural number So, P(k+1) is true, whenever P(k) is true. Hence, P(n) is true.
Discussion
7.

Using the principle of mathematical induction, prove each of the following for all `n in N` `3^(n) ge 2^(n)`

Answer» Clearly, ` 3^(1) ge 2^(1)`. So, the result is true for n = 1.
Let it be true for n = k. Then,
` 3^(k) gt 2^(k) and 3 gt 2 rArr 3^(k) * 3 ge 2^(k) * 2 rArr 3^(k+1) ge 2 ^(k+1)`.
So, whenever the result is true for k, then it is also true for (k+1).
Discussion
8.

`1+3+3^(2)+…….+3^(n-1) =((3^(n)-1))/(2)`

Answer» Let `P (n) :1+3+3^(2)+…..+3^(n-1)=(3^(n)-1)/(2)`
`" If "n =1 , "the " L.H.S. =1`
`R.H.S. =(3^(1)-1)/(2)=(3-1)/(2)=1`
`:. " "L.H.S. =R.H.S.`
Therefore the statement P (n) is true for n=1
Let P (n) be true for n=K.
`:. P (k) : 1+3+3^(2) +.....+3^(k-1) =(3^(K)-1)/(2)`
`P (k+1) :1+3+3^(2)+......+3^(k)`
`=1+3+3^(2)+......+3^(k)`
`=1+3+3^(2)+.......+3^(k-1)+3^(k)`
`=(3^(k)-1)/(2) +3^(k)`
`=(3^(k)-1+2.3^(k))/(2)=((1+2)3^(k)-1)/(2)`
`=(3.3^(k)-1)/(2)=(3^(k+1)-1)/(2)`
Then the statement P (n) is also true for n=K +1 ,
Hence form the principle of mathematical induction P (n) is true for `n in N`
Discussion
9.

Prove that `1/(n+1)+1/(n+2)+...+1/(2n)> 13/24` ,for all natural number `n>1`.

Answer» P(n) : `(1)/(n+1)+(1)/(n+2)+ . . .+(1)/(2n)gt(13)/(24)`, for all natural numbers `ngt1`.
Step I We observe that, P(2) is true,
`P(2):(1)/(2+1)+(1)/(2+2)gt(13)/(24).`
`(1)/(3)+(1)/(4)gt(13)/(24)`
`(4+3)/(12)gt(13)/(24)`
`(7)/(12)gt(13)/(24)` which is true
Step II Now, we assume that P(n) is true,
For n=k,
`P(k):(1)/(k+1)+(1)/(k+2)+ . . . +(1)/(2k)gt(13)/(24).`
Step III Now, to prove P(k+1) true we have to show that
`P(k+1):(1)/(k+1)+(1)/(k+2)+ . . . +(1)/(2k)+(1)/(2(k+1))gt(13)/(24)`
Given. `(1)/(k+1)+(1)/(k+2)+ . . . +(1)/(2k)gt(13)/(24)`
`(1)/(k+1)+(1)/(k+2)+ . . . +(1)/(2k)+(1)/(2(k+1))gt(13)/(24)+(1)/(2(k+1))`
`(13)/(24)+(1)/(2(k+1))gt(13)/(24)`
`because(1)/(k+1)+(1)/(k+2)+ . . . +(1)/(2k)+(1)/(2(k+1))gt(13)/(24)`
So, P(k+1) is true, whenever p(k) is true. Hence, P(n) is true.
Discussion
10.

`1+1/(1+2)+1/(1+2+3)+1/(1+2+3+n)=(2n)/(n+1)`

Answer» Let P (n) `:1 +(1)/(1+2)+(1)/(1+2+3)+……`
`+(1)/((1+2+3+.....+n))=(2n)/(n+1)` ltbr. For n=1
`L.H.S. =1`
`R.H.S. =(2.1)/(1+1) =(2)/(2) =1`
`:. " "L.H.S.=R.H.S.`
Therefore,P (n) is true for n=1
Let P (n) true for n =K .
`P(k) :1 (1)/(1+2) +(1)/(1+2+3)+....`
`+(1)/(1+2+3+....+K)=(2K)/(K+1)`
For n =k +1
`P (k+1) =1 + (1)/(1+2)+(1)/(1+2+3)+.....+(1)/(1+2+3+.....+K)`
`+(1)/(1+2+3+.....+K+(K+1))`
`(2K)/(K+1)+(1)/(1+2+3+......+K+(K+1))`
`=(2k)/(K+1) +(1)/((K+1)(K+2))=(2K(K+2)+2)/((K+1)(K+2))`
`=(2(K^(2)+2K+1))/((k+1)(K+2)) =(2(K+1)^(2))/((K+1)(K+2))`
`(2(K+1))/(K+2)=(2(K+1))/((K+1)+1)`
`rArr` P (n) is also true for n=k+1
hence form the principle of mathematical induction P (n) is true for all natural numbers n.
Discussion
11.

Let `U_1 = 1, U_2=1 and U_(n+2)=U_(n+1)+U_n` for `n>=1`. Use mathematical induction to such that : `U_n=1/(sqrt(5)){((1+sqrt(5))/2)^n-((1-sqrt(5))/2)^n}` for all `n>=1`.

Answer» Given `u_(1) = 1, u_(2) = 1,u_(n+2) = u_(n+1) + u_(n) n ge 1`
`u_(n) = 1/(sqrt(5))[((1+sqrt(5))/(2))^(n) - ((1-sqrt(5))/(2))^(n) ]`
`:. u_(2) = (1)/(sqrt(5))[((1+sqrt(5))/(2))^(2)-((1-sqrt(5))/(2))^(2)]`
`= (1)/(sqrt(5))((1+sqrt(5))/(2)+(1-sqrt(5))/(2))((1+sqrt(5))/(2)-(1-sqrt(5))/(2))`
`= (1)/(sqrt(5)) (1 xx sqrt(5)) = 1`
Thus m `u_(n)` is true for `n = 2`
Let it be true for `n = k gt 2`" "(as given `u_(2) = 1`). Then
`u_(k) = (1)/(sqrt(5))[((1+sqrt(5))/(2))^(k) - ((1-sqrt(5))/(2))^(k)]`
Consider that ` u_(k+1) = u_(k) + u_(k-1)` [Using `u_(n+2) = u_(n+1) + u_(n)`] ltbr gt `:. u_(k+1) = 1/(sqrt(5)) [((1+sqrt(5))/(2))^(k) - ((1-sqrt(5))/(2))^(k)]`
`+ (1)/(sqrt(5)) [((1+sqrt(5))/(2))^(k-1)-((1-sqrt(5))/(2))^(k-1)]`
`= (1)/(sqrt(5)) [((1+sqrt(5))/(2))^(k) + ((1+sqrt(5))/(2))^(k-1)]`
`- [((1-sqrt(5))/(2))^(k) + ((1+sqrt(5))/(2))^(k-1)]`
`= 1/(sqrt(5)) [((1+sqrt(5))/(2))^(k-1){(1+sqrt(5))/(2) +1}]`
`- [((1-sqrt(5))/(2))^(k-1){(1-sqrt(5))/(2)+1}]`
`= (1)/(sqrt(5)) [((1+sqrt(5))/(2))^(k-1){(6+2sqrt(5))/(4)}-((1-sqrt(5))/(2))^(k-1){(6-2sqrt(5))/(4)}]`
`= 1/(sqrt(5)) [((1+sqrt(5))/(2))^(k-1)((sqrt(5)+1)/(2))^(2)-((1-sqrt(5))/(2))^(k-1)((sqrt(5)-1)/(2))^(2)]`
`= (1)/(sqrt(5))[((1+sqrt(5))/(2))^(k+1)-((1-sqrt(5))/(2))^(k+1)]`
So, `u_(k+1)` is also true.
Hence, by principle of mathematical induction `u_(n)` is true `AA n gt 1`.
Discussion
12.

Prove the following by the principle ofmathematical induction:`1/(2. 5)+1/(5. 8)+1/(8. 11)++1/((3n-1)(3n+2))=n/(6n+4)`

Answer» Let
`P (n) : (1)/(2.5)+(1)/(5.8)+(1)/(8.11)+…..+(1)/((3n-1)(3n+2))`
`=(n)/((6n+4))`
for n=1
`L.H.S. =(1)/(2.5)+(1)/(10)`
and `R.H.S. =(1)/(6.1+4) =(1)/(6+4)=(1)/(10)`
`rArr " "L.H.S. =R.H.S.`
Therefore given statement is true for n=1
Let the statement P (n) be true for n=k
`:. P(k) =(1)/(2.5)+(1)/(5.8)+(1)/(8.11)+......`
`+(1)/((3k-1)(3k+2))=(k)/(6k+4)`
For n=K+1
`P(K+1) : (1)/(2.5)+(1)/(5,8)+(1)/(8.11) +.......`
`+(1)/[[(3k+1)-1][3(k+1)+2)]`
`=(1)/(2.5)+(1)/(5.8)+(1)/(8.11)+.....+(1)/((3k-1)(3k+2))`
`+(1)/((3k+2)(3k+5))`
`((1)/(2.5)+(1)/(5.8)+(1)/(8.11)+.........+(1)/((3k-1)(3k+2)))`
`+(1)/((3k+2)(3k+5))`
`=(k)/(6k+4)+(1)/((3k+2)(3k+5))`
`=(1)/((3k+2))((K)/(2)+(1)/(3k+5))`
`=(1)/(3k+2)[(3k^(2)+5k+2)/(6k+10)]`
`=(1)/(3k+2)[(3k^(2)+3k+2K+2)/(6k+10]]`
`=(1)/(3k+2).[(3k(K+1)+2(k+1))/(6k+10)]`
`=(1)/(3k+2).((3k+2)(k+1))/(6k+10)=(k+1)/(6k+10)`
Then given statment P (n) is also true for n=K+1
Hence given statement P (n) is true for all values of n where `n in N`
Discussion
13.

prove that `2+4+6+…2n=n^(2)+n`, for all natural numbers n.

Answer» Let `P(n): 2+4+6+ . . +2n=n^(2)+n`
For all natural number n.
Step I We observe that P(1) is true.
`P(1):2=1^(2)+1`
2=2, which is true.
Step II Now, assume that P(n) is true for n=k.
`:.P(k):2+4+6+ . . .+2k=k^(2)+k`
Step II To prove that `P(k+1):2+4+6+8 . . . +2k+2(k+1)`
`=k^(2)+k+2(k+1)`
`=k^(2)+2k+1+k1`
`=(k+1)^(2)+k+1`
So, P(k+1) is true, whenever P(k) is true.
Hence, P(n) is true.
Discussion
14.

Let `A_(n) = a_(1) + a_(2) + "……" + a_(n), B_(n) = b_(1) + b_(2) + b_(3) + "…." + b_(n), D_(n) = c_(1) + c_(2) + "….." + c_(n)` and `c_(n) = a_(1)b_(n) + a_(2)b_(n-1) + "……." + a_(n)b_(1)Aan in N`. Using mathematical induction , prove that (a) `D_(n) = a_(1)B_(n) + a_(2)B_(n-1) + "....."+a_(n)B_(1) = b_(1)A_(n) + b_(2)A_(n-1) + "......"+b_(n)A_(1) AA n in N` (b) `D_(1) + D_(2) + "......"+ D_(n) = A_(1)B_(n) + A_(2)B_(n-1) + "....." + A_(n)B_(1) AA n in N`

Answer» Let ` P(n) : D_(n) a_(n)B_(n) + a_(n)B_(n-1)+"……."+a_(n)B_(1)`
For `n = 1, P(1) = D_(1) = a_(1)B_(1) = a_(2)b_(1) = c_(1)`
Thus, ` P(1)` is true.
Let `P(n)` be true for `n = k`
i.e., `P(k) = D_(k) = a_(1)B_(k)= a_(1)B_(k) + a_(2)B_(k-1) + "....."+ a_(k)B_(1)`
Now, `D_(k+1) = D_(k) + c_(k+1)`
`= a_(1)B_(k) + a_(2)B_(k-1) + "......"+a_(k)B_(1) +a_(1)b_(k+1) + a_(2)b_(k) + "....."+a_(k+1)b_(1)`
`= a_(1)(B_(k)+b_(k+1))+ a_(2)(B_(k-1)+b_(k)) + "......" + a_(k)(B_(1) + b_(2)) + a_(k+1)b_(1)` ltbrlt `= a_(1)B_(k+1)+a_(2)B_(2)+"......."+a_(k)B_(2) +a_(k+1)B_(1)`.
Thus, `P(k+1)` is also true whenever `P(k)` is true.
So, by the principle of mathematical induction `P(n)` is true
for any natural number n.
Let `P_(1)(n) : D_(n) = b_(1)A_(n) + b_(2)A_(n-1)+"......."+b_(n)A_(1)`
For `n = 1, P_(1)(1) = D_(1) = b_(1)A_(1) = b_(1)a_(1) = c_(1)`
Thus, `P_(1)(1)` is true.
Let, `P_(1)(n)` be true for `n = k`.
i.e, `P_(1)(k) = b_(1)A_(k) + b_(2)A_(k-1) + "....." + b_(k)A_(1)`
Now, `D_(k+1) = D_(k) +c_(k+1)`
`= b_(1)A_(k) + b_(2)a_(k-1) + "......." + b_(k)A_(1) + a_(1)b_(k+1) + a_(2)b_(k) + "....." + a_(k+1)b_(1)`
`= b_(1)(A_(k) + a_(k+1))b_(2)(A_(k-1)+a_(k))+"......"+b_(k)(A_(1) + a_(2)) + b_(k+1)a_(1)`
`= b_(1)A_(k+1) + b_(2)A_(k) + "......"+b_(k)A_(2) + b_(k+1)A_(1)`
Thus, `P_(1)(k+1)` is also true.
So, by the principle of mathematical induction `P_(1)(n)` is true for any natural number n.
(b) Say, `P_(2)(n) = D_(1) + D_(2) + "......"+D_(n) = A_(1)B_(n) + A_(2)B_(n-1) + "......" + A_(n)B_(2)`
For `n = 1, P_(2)(1) = D_(1) = A_(1)B_(1) = a_(1)b_(1) = c_(1)`
Thus, `P_(2)(1)` is true .
Let `P_(2)(n)` be true for `n = k`
i.e, ` P_(2)(k) = A_(1)B_(k) + A_(2)B_(k+1) + "......" + A_(k)B_(1)`
Now, `P_(2)(k+1)`
`= P_(2)(k) + D_(k+1)`
`= A_(1)B_(k) + A_(2)B_(k-1) +"........"+)A_(k)B_(1) +b_(1)A_(k+1) + b_(2)A_(k) + "......" + b_(k)A_(2) + b_(k+1)A_(1)`
`= A_(1)(B_(k) + b_(k+1)) + A_(2)(B_(k-1)+b_(k)) + "....." + A_(k)(B_(1) + b_(2)) + A_(k+1)b_(1)`
` = A_(1)B_(k+1) + A_(2)B_(k) + "....." + A_(k)B_(2) + A_(k+1)B_(1)`
Thus, `P_(2)(k+1)` is also true.
So, by the principle of mathematical induction, `P_(2)(n)` is true for any natural number n.
Discussion
15.

A sequence `a_(1),a_(2),a_(3), . . .` is defined by letting `a_(1)=3` and `a_(k)=7a_(k-1)`, for all natural numbers `k≥2`. Show that `a_(n)=3*7^(n-1)` for natural numbers.

Answer» A sequence `a_(1),a_(2),a_(3), . . .` is defined by letting `a_(1)=3` and `a_(k)=7a_(k-1)`, for all natural numbers `kle2`.
Let `P(n):a_(n)=3*7^(n-1)` for all natural numbers.
Step I We observe P(2) is true.
For n=2, `a_(2)=3*7^(2-1)=3*7^(1)=21` is true.
As `a_(1)=3,a_(k)=7a_(k-1)`
`rArra_(2)=7*a_(2-1)=7*a_(1)`
`rArra_(2)7xx3=21 [becausea_(1)=3]`
Step II Now, assume that P(n) is true for n=k.
`P(k):a_(k)=3*7^(k-1)`
Step III Now, to prove P(k+1) is true, we have to show that
`P(k+1):a_(k+1)=3*7^(k+1-1)`
`a_(k+1)=7*a_(k+1-1)+7*a_(k)`
`=7*3*7^(k-1)=3*7^(k-+1)`
So, P(k+1) is true, whenever p(k) is true. Hence, P(n) is true.
Discussion
16.

Prove that`1+2+2^(2)+ . . .+2^(n)=2^(n+1)-1`, for all natural number n.

Answer» Consider the given statement
`P(n):1+2+2^(2)+ . . .2^(n)=2^(n+1)-1`, for natural numbers n.
Step I We observe that P(0) is true.
`P(1):1=2^(0+1)-1`
`1=2^(1)-1`
1=2-1
1=1, which is true.
Step II Now, assume that P(n) is true for n=k.
So, P(k) : `1+2+2^(2)+ . . .2^(k)=2^(k+1)-1` is true.
Step III Now, to prove P(k+1) is true.
`P(k+1):1+2+2^(2)+. . .+2^(k)+2^(k+1)`
`=2^(k+1)-1+2^(k+1)`
`=2*2^(k+1)-1`
`=2^(k+1)+1-1`
So, P(k+1) is true, whenever P(k) is true.
Hence, P(n) is true.
Discussion
17.

prove that 1+5+9+ . . .+(4n-3)=n(2n-1), for all natural number n.

Answer» Let P(n):1+5+9+ . . .+(4n-3)=n(2n-1), for all natural number n.
Step I We observe that P(1) is true.
`P(1):1=1(2xx1-1), 1=2-1` and 1=1, which is true.
Step II Now assume that P(n) is true for n=k.
So, P(k):1+5+9+ . . .+(4k-3) = k(2k-1) is true.
Step III Now, to prove P(k+1) is true.
(P(k+1):1+5+9+. . . +(4k-3)+4k+1)-3
=k(2k-1)+4(k+1)-3
`=2k^(2)-k+4k+4-3`
`=2k^(2)+3k+1`
`=2k^(2)+2k+k+1`
2K(k+1)+1(k+1)
=(k+1)(2k+1)
=(k+1)[2k+1+1-1]
=(k+1)[2(k+1)-1]
So, P(k+1) is true, whenever p(k) is true, hence p(n) is true.
Discussion
18.

A sequence `a_(1),a_(2),a_(3), . . .` is defined by letting `a_(1)=3` and `a_(k)=7a_(k-1)`, for all natural numbers `kle2`. Show that `a_(n)=3*7^(n-1)` for natural numbers.

Answer» We have a sequence `a_(1),a_(2),a_(3)"…."` is defined by letting `a_(1) = 3` and `a_(k) = 7a_(k-1)`, for all natural number `k ge 2`.
Let `P(n) : a_(n) = 3 xx 7^(2-1) = 3 xx 7^(1) = 21`
Also, `a_(1) = 3, a_(k) = 7a_(k-1)`
`rArr a_(2) = 7a_(1) =a = 7 xx 3 = 21`
Thus, `P(2)` is true.
Now, assume that `P(k)` is true.
`:. a_(k)= 3 xx 7^(k-1)`
Now, to prove `P(k+1)` we have to show that
`a_(k+1) = 3 xx 7^(k+1-1)`
Given that `a_(k) = 7a_(k-1)`
So, `a_(k+1) = 7a_(k+1-1)`
`= 7a_(k)`
`= 7 xx 3 xx 7^(k-1)`
`= 3 xx 7^((k+1)-1)`
Hence, `P(k+1)` is true whenever `P(k)` is true.
So, by the principle of mathematical inducton, `P(n)` is true for any natural number `n`.
Discussion
19.

A sequence `d_(1),d_(2),d_(3) . . .` is defined by letting `d_(1)=2` and `d_(k)=(d_(k-1))/(k),` for all natural numbers, `k≥2`. Show that `d_(n)=(2)/(n!)`, for all `n in N`.

Answer» Let P(n) : `d_(n)=(2)/(n!)AAninN`, to prove P(2) is true.
Step I `P(2):d_(2)=(2)/(2!)=(2)/(2xx1)=1`
As, given `d_(1)=2`
`rArrd_(k)=(d_(k-1))/(k)`
`rArrd_(2)=(d_(1))/(2)=(2)/(2)=1`
Hence, P(2) is true
Step II Now, assume that P(k) is true.
`P(k):d_(k)=(2)/(k!)`
Step III Now, to prove that P(k+1) is true, we have to show that `P(k+1):d_(k+1)=(2)/(k+1)!`
`d_(k+1)=(d_(k+1-1))/(k)=(d_(k))/(k)`
`=(2)/(k!k)=(2)/(k+1)!`
So, P(k+1) is true. Hence, P(n) is true.
Discussion
20.

Prove by the principle of mathematical induction that for all `n in N`:`1+4+7+...+(3n-2)=1/2n(3n-1)`

Answer» Let the given statement be P(n). Then,
`P(n) : 1+4+7+10+…+(3n-2) = 1/2n(3n - 1)`.
Putting n = 1 in the given statement, we get
LHS = 1 and RHS ` = 1/2 xx 1 xx (3 xx 1-1) = 1`.
`:. ` LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
`P(k) : 1+4+7+10 +...+(3k-2)= 1/2 k(3k-1)`. ....(i)
Now, `1+4+7+...+(3k-2)+{3(k+1)-2}`
` = {1+4+7+...(3k-2)}+(3k+1)`
` =1/2 k(3k-1)+(3k+1)" "` [using (i)]
` = 1/2 (3k^(2)-k+6k+2)= 1/2 (3k^(2)+5k+2)`
` = 1/2 (3k^(2)+3k+2k+2)= 1/2 * {3k(k+1)+2(k+1)}`
` = 1/2 (k=1)(3k+2)= 1/2 (k+1) {3(k+1)-1}`.
`:. P(k+1):1+4+7+...+{3(k+1)-2}= 1/2 (k+1){3(k+1)-1}`.
This shows that P(k+1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k+1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, we have
` 1+4+7+...+(3n-1)= 1/2 n(3n-1)" for all " n in N`.
Discussion
21.

Prove the following by using the principle of mathematical induction for all `n in N`:`1^3+2^3+3^3+...+n^3=((n(n+1))/2)^2`

Answer» Let the given statement be P(n). Then,
` P(n) : 1^(3)+2^(3)+3^(3)+…+n^(3)= {(n(n+1))/2}^(2)`.
Putting n = 1 in the given statement, we get
LHS = `1^(3) = 1 and RHS = ((1xx2)/2)^(2) = 1^(2) = 1`.
`:. ` LHS = RHS .
Thus, P(1) is true.
Let P(k) be for some `k in N`. Then,
`P(k) : 1^(3)+2^(3)+3^(3) + ... k^(3) = {(k(k+1)^(2))/2}^(2)." "` ...(i)
Now, `1^(3)+2^(3)+3^(3)+...+k^(3)+(k+1)^(3)`
` = {1^(3)+2^(3)+3^(3)+...+k^(3)}+(k+1)^(3)`
` = {(k(k+1))/2}^(2) +(k+1)^(3)" "` [using (i)]
`=(k+1)^(2){k^(2)/4+(k+1)}=(k+1)^(2){(k^(2)+4k+4)/4}`
`=((k+1)^(2)(k+2)^(2))/4 = {((k+1){(k+1)+1})/2}^(2)`
` :. " " P(k+1): 1^(3)+2^(3)+3^(3)+...+(k+1)^(3)=[((k+1){(k+1)+1})/2]^(2)`.
This shows that P(k+1) is true, whenever P(k) is true.
`:." " P(1) ` is true and P(k+1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, we have
` 1^(3)+2^(3)+3^(3)+...+n^(3)={(n(n+1))/2}^(2)" for all " n in N`.
Discussion
22.

A sequence `b_(0),b_(1),b_(2), . . .` is defined by letting `b_(0)=5` and `b_(k)=4+b_(k-1)`, for all natural number k. Show that `b_(n)=5+4n`, for all natural number n using mathematical induction.

Answer» Consider the given statement,
`P(n):b_(n)=5+4n`, for natural numbers given that `b_(0)=5` and `b_(k)=4+b_(k-1)`
Step I P(1) is true
`P(1):b_(1)=5+4xx1=9`
As `b_(0)=5,b_(1)=4+b_(0)=4+5=0`
Hence, P(1) true.
Step II Now, assume that P(n) true for n=k.
`P(k):b_(k)=5+4k`
Step III Now, to prove P(k+1) is true, we have to show that
`:. P(k+1):b_(k=1)=5+4(k+1)`
`b_(k+1)=4+b_(k+1-1)`
`=4+b_(k)`
`=4+5+4k=5+4(k+1)`
So, by the mathematical induction P(k+1) is true whenever p(k) is ture, hence P(n) is true.
Discussion
23.

prove that `2nlt(n+2)!` for all natural numbers n.

Answer» Consider the statement
`P(n):2nlt(n+2)!` for all natural number n.
Step I We observe that, P(1) is true `P(1):2(1)lt(1+2)!`
`rArr 2lt3!rArr2lt3xx2xx1rArr2lt6`
Hence, P(1) is true.
Step II Now, assume that p(n) is true for n=k,
`P(k):2klt(k+2)!` is true
Step III Now, assume that P(n) is true for n=k,
`P(k):2klt(k+2)!` is true.
Step III To prove P(k+1) is true, we have to show that
`P(k+1):2(k+1)lt(k+1+2)!`
Now, `2klt(k+2)!`
`2k+2lt(k+2)!`
`2k+2ltIk+2)!+2`
`2(k+1)lt(k+2)!+2` . . .(i)
Also, `(k+2)!+2lt(k+3)!` . . . (ii)
From Eqs. (i) and (ii),
`2(k+1)lt(k+1+2)!`
So, P(k+1) is true, whenever P(k) is true.
Hence, by principle of mathematical induction P(n) is true.
Discussion
24.

prove that `n^(2)lt2^(n)`, for all natural number `n≥5`.

Answer» Consider the given statement
`P(n):n^(2)ltn^(n)` for all natural number `nle5`.
Step I We observe that P(5) is true `P(5):5^(2)lt2^(5)`
`=25lt32`
Hence, P(5) is ture.
Step II Now, assume that P(n) true for n=k.
`P(k)=k^(2)lt2^(k)` is true.
Step III Now, to prove P(k+1) is true, we have to show that
`P(k+1):(k+1)^(2)lt2^(k+1)`
Now, `k^(2)lt2^(k)=k^(2)+2k+1lt2^(k)+2k+1` . . . (i)
`=(k+1)^(2)lt2^(k)+2k+1`
Now, `(2k+1)lt2^(k) =2^(k)+2k+2lt2^(k)+2^(k)`
`=2^(k)+2k+1lt2*2^(k)`
`=2^(k)+2k+1lt2^(k=1)` . . . (ii)
Form Eqs. (i) and (ii), we get `(k+1)^(2)lt2^(k+1)`
So, P(k+1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true for all natural numbers `nle5`.
Discussion
25.

prove that `n(n^(2)+5)` is divisible by 6, for each natural number n.

Answer» Let `(P):n(n^(n)+5)` is divisible by 6, for each natural number n.
Step I We observe that P(1) is true.
`P(1):1(1^(2)+5)=6`, which is divisible by 6.
Step II Now, assume that P(n) is true for n=k.
`P(k):k(k^(2)+5)` is divisible by 6.
`:.k(k^(2)+5)=6q`
Step III Now, to prove p(k+1) is true, we have
`P(k+1) : (k+1) [(k+1)^(2)+5]`
`=(k+1)[k^(2)+2k+1+5]`
`=(k+1)[k^(2)+2k+1+5]`
`=(k+1)[k^(2)+2k+6]`
`=k^(3)+2k^(2)+6k+k^(2)2k+6`
`=k^(3)+3k^(2)+8k+6`
`=k^(3)+5k+3k^(2)+3k+6`
`=k(k^(2)+5)+3(k^(2)+k+2)`
`=(6q)+3(k^(2)+k=2)`
We know that, `k^(2)+k+2` is divisible by 2, where, k is even or odd.
Since, `P(k+1):6q+3(k^(2)+k+2)` is divisible by 6. So, P(k+1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true.
Discussion
26.

If n be any natural number then by whichlargest number `(n^3-n)`is always divisible?3(b) 6 (c) 12 (d) 18

Answer» Let P(n) : `n^(3)-n` is divisible by 6, for each natural bumber `nle2`.
Stwp I We observe that P(2) is true. P(2) : `(2)^(3)-2`.
`rArr 8-2=6`, which is divisible by 6.
Stwep II Now, assume that P(n) is true for n=k.
P(k) : `k^(3)-k` is divisible by 6.
`:.k^(3)-k=6q`
Step III To prove P(k+1) is true
`P(k+1):(k+1)^(3)-(k+1)`.
`=k^(3)+1+3k(k+1)-(k+1)`
`=k^(3)+1+3k^(2)+3k-k-1`
`=k^(3)-k+3k^(2)+3k`
`=6q+3k(k+1)` [from step II]
We know that, 3k (k+1) is divisible by 6 for each natural number n=k.
So, P(k+1) is true. Hence, by the principle of mathematical induction P(n) is true.
Discussion
27.

Prove that for any natural numbers n, `7^(n)-2^(n)` is divisible by 5.

Answer» Consider the given statement is
`P(n):7^(n)-2^(n)` is
Step I We observe that P(1) is true.
`P(1)=7^(1)-2^(1)=5`, which is disivible by 5.
Step II Now, assume that P(n) is true for n=k.
`P(k)=7^(k)-2^(k)=5q`
Step III Now, to prove P(k+1) is true,
`P(k+1):7^(k+1)-2^(k+1)`.
`=7^(k)*2^(k)*2`
`=7^(k)*(5+2)-2^(k)*2`
`=7^(k)*5+2*7^(k)-2^(k)*2`
`5*7^(k)+2(7^(k)-2^(k))`
`=5*7^(k)+2(5q)`
`=5(7^(k)+2q)`, which is divisible by 5. [from step II]
So, P(k+1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true for any natural number n.
Discussion
28.

Prove the following by using the principle of mathematical induction for all `n in N`:`1^3+2^3+3^3+dotdotdot+n^3=((n(n+1))/2)^2`

Answer» `P(n) : 1^(3) + 2^(3) + 3^(3) + "…. + n^(3) = (n^(2)(n+1)^(2))/(4)`
For `n = 1`,
L.H.S. `= 1^(3) = 1` and `R.H.S. = (1^(2).2^(2))/(4) = 1`
Thus, `P(1)` is true.
Let `P(n)` be true for some `n = k`.
i.e, `1^(3) + 2^(3) + 3^(3) + "...." + k^(3) = (k^(2) (k+1)^(2))/(4) "....."(1)`
Now, we have to prove that ` P(n)` is true for ` n = k + 1`.
i.e, `1^(3) + 2^(3) + 3^(3) + ".... " +k^(3) + (k+1)^(3) = ((k+1)^(2)(k+2)^(2))/(4)`
Adding `(k+1)^(3)` on both sides of `(1)`, we get `1^(3) + 2^(3) + 3^(3) + ".... " +k^(3) + (k+1)^(3)`
`= (k^(2)(k+1)^(2))/(4)+(k+1)^(3)`
`= ((k+1)^(2))/(4) (k^(2)+4(k+1))`
`= ((k+1)^(2)(k+2)^(2))/(4)`
Thus, `P (k+1)` is true whenever `P(k)` is true.
Hence, by the principle of malthematical induction, statement `P(n)` is true for all natural numbers.
Discussion
29.

Using the principle ofmathematical induction prove that`1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)++1/(1+2+3++n)=(2n)/(n+1)`for all `n in N`

Answer» `P(n) : 1 + (1)/(1+2) + (1)/(1+2+3) + "…."+(1)/(1+2+3+"…."+n) = (2n)/(n+1)`
For `n = 1`,
` L.H.S. = 1` and `R.H.S. = (2 xx 1)/(1+1) = 2/2 = 1`
Thus, ` P(1)` is true.
Let `P(n)` be true for some `n = k`.
i.e, `1+(1)/(1+2)+"……"+(1)/(1+2+3)+"......"+(1)/(1+2+3+"...."k) = (2k)/(k+1)"....."(1)`
Now, we have to prove that `P(n)` is true for `n = k + 1`.
ie., `1+(1)/(1+2) +"..."+(1)/(1+2+3)+"..."+(1)/(1+2+3+"....."+(k+1))`
`= (2(k+1))/(k+2)`
Adding `(1)/(1+2+3+"......"+(k+1))` on both sides of `(1)`, we get
`1+(1)/(1+2)+"...."+(1)/(1+2+3) +"...."(1)/(1+2+3+"....."+k) + (1)/(1+2+3+"...."+(k+1))`
` = (2k)/(k+1)+(1)/(1+2+3+"....."+(k+1))`
`= (2k)/(k+1)+(1)/(((k+1)(k+2))/(2))`
[Using `1+2+3+"...."+n = (n(n+1))/(2)`]
`= (2k)/((k+1)) + (2)/((k+1)(k+2))`
`= (2)/(k+1)((k(k+2)+1)/(k+2))`
`= (2(k+1)^(2))/((k+1)(k+2))`
`= (2(k+1))/((k+2))`
Thus, `P(k+1)` is the whenever `P(k)` is true.
Hence by the principle of mathemetical induction, statement `P(n)` is true for all natural numbers.
Discussion
30.

prove that `3^(2n)-1` is divisible by 8, for all natural numbers n.

Answer» Let `P(n):3^(2n)-1` is divisible by 8, for all natural numbers.
Step I We observe that P(1) is true.
`P(1):3^(2(1))-1=3^(2)-1`
=9-1=8,which is divisible by 8.
Step II Now, assume that P(n) is true for n=k.
`P(k):3^(2k)-1=8q`
Step III Now, to prove P(k+1) is true.
`P(k+1):3^(2(k+1))-1`
`=3^(2k)*3^(2)-1`
`=3^(2k)*(8+1)-1`
`=8*3^(2k)+3^(2k)-1`
`=8*3^(2k)+8q`
`=8(3^(2k)+q)` [from step II]
Hence, P(k+1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.
Discussion
31.

`7^(2n)+2^(3n-3)*3^(n-1)` is divisible by 25

Answer» by appliying the mathematical induction formula,
we have, `(7^2)^n + (2^3)^(n-1) * 3^(n-1)`
`(49)^n + (8)^(n-1).3^(n-1)`
`49^n + 24^(n-1)`
`(50-1)^n + (25-1)^(n-1)`
`.^nC_0*(50)^n - .^nC_1(50)^(n-1) .....................- .^(n-1)C_(n-1)*(-1)^(n-1) `
`25k + (-1)^n + (-1)^(n-1)`
now, if n= odd then n-1 = evenand if n= even then n-1 = oddtherefore the term ` underbrace((-1)^n + (-1)^(n-1) )` is always going to be zero (0)therefore, the equation will be as = 25kso, it is divisible by 25.
Discussion
32.

Using the principle of mathmatical induction, prove each of the following for all `n in N` `(1+1/1)(1+1/2)(1+1/3)...(1+1/n)=(n+1)`.

Answer» `P(k): (1+1)(1+1/2)(1+1/3)…(1+1/k)= (k+1)`.
Now, `{(1+1)(1+1/2)(1+1/3)…(1+1/k)}(1+1/(k+1))`
` = (k+1)(1+1/(k+1)) = (k+1)*((k+2))/((k+1)) = (k+2)`.
Discussion
33.

`3^(2n+2)-8n-9` divisible by `8`

Answer» Let `P(n) =3^(2n+2) -8n-9`
for n=1
`p(1) =3^(4)-8(1) -9 =81 -17 =64 =8 (8)`
Which is divisible by 8
`rArrP (n)` be true for n=1
Let P (n) be true for n=K.
`:.P(k) : 3^(2K+2) -8k-9=8lambda (" say ")`
`" Where " lambda in I`
for n=k+1
`P(k+1):3^(2(k+1)+2) -8(k+1)-9`
`=3^(2).3^(2k+2) -8k-8-9`
`=9[8lambda+8k+9]-8k-17`
`=9.8lambda+72k+81-8K-17`
`=9.8lambda+64lambda+64`
`=8[9lambda+8k+8]`
which is divisible by 8
`rArr` P (n) is also true for n=K+1
Hence from the principle of mathematical induction p(n) is true for all natural numbes n.
Discussion
34.

Using principle of mathematical induction, prove the following `1+3+5+...+(2n - 1)=n^2`

Answer» Let the given statement be P (n). Then,
` P(n) : 1+3+5+7+…+(2n-1)= n^(2) `
Putting n = 1 in the given statement, we get
LHS = 1 and RHS = ` 1^(2) = 1`.
`:. ` LHS = RHS.
Thus, P(1) is true.
Let P (k) be true. Then,
`P(k) : 1+3+5+7+...+(2k-1) = k^(2)`. ....(i)
Now, ` 1+3+5+7+...+(2k-1)+{2(k+1)-1}`
` = {1+3+5+7+...+(2k-1)}+(2k+1)`
` =k^(2) (2k+1)" "` [using (i)]
` = (k+1)^(2)`
` :. P(k+1): 1+3+5+7+...(2k-1)+{2(k+1)-1}=(k+1)^(2)`
This shows that P(k+1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k+1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, we have
`1+3+5+7+...+(2n-1)= n^(2) " for all " n in N`.
Discussion
35.

Prove the following by using the principle of mathematical induction for all `n in N`:`(1+3/1)(1+5/4)(1+7/9)...(1+((2n+1))/(n^2))=(n+1)^2`

Answer» `P(k):(1+3)(1+5/4)(1+7/9)…{1+((2k+1))/(k^(2))}=(k+1)^(2)`.
Now, `(1+3)(1+5/4)…(1+(2k+1)/k^(2)){1+(2(k+1)+1)/((k+1)^(2))}`
` = (k+1)^(2) xx ((k+1)^(2)+(2k+3))/((k+1)^(2)) = (k^(2)+3k+4) = (k+2)^(2)`.
Discussion
36.

For all n `inN,3*5^(2n+1)+2^(3n+1)` is divisible by(A) 19 (B) 17 (C) 23 (D) 25A. 19B. 17C. 23D. 25

Answer» Correct Answer - B::C
Given that `3*5^(2n+1)+2^(3n+1)`
For n=1,
`3*5^(2(1)+1)+2^(3(1)+1`
`=3*5^(3)+2^(4)`
`=3xx125+16=375+16=391`
Now, `391=17xx23`
which is divisible by both 17 and 23.
Discussion
37.

Using the principle of mathematical induction prove that `:``1. 3+2. 3^2+3. 3^3++n .3^n=((2n+1)3^(n+1)+3)/4^`for all `n in N`.

Answer» Let the given statement be P(n) . Then,
`P(n) : 1*3+2*3^(2)+3*3^(3)+…+n*3^(n) = ((2n-1)3^(n+1)+3)/4` .
Putting n = 1 in the given statement, we get
LHS ` = 1*3 and RHS = ((2XX1-1)3^(1+1)+3)/4=(1xx9+3)/4 = 12/4 = 3`.
`:. ` LHS = RHS.
Thus , P(1) is true.
Let P(k) be true. Then,
` P(k) : 1*3+2*3^(2)+3*3^(3)+...+k*3^(k) = ((2k-1)3^(k+1)+3)/4 `. ...(i)
` :. 1*3+2*3^(2)+3*3^(30+...+k*3^(k)+(k+1) 3^((k+1))`
`=((2k-1)3^(k+1)+3)/4 +(k+1) 3^(k+1)" " ` [using (i)]
`=((2k-1)3^(k+1)+3+4(k+1)3^(k+1))/4 = ((2k-1+4k+4)3^(k+1)+3)/4 `
` = ((6K+3) 3^(k+1)+3)/4 = ({2(k+1)-1}3^((k+1)+1)+3)/4 ` .
` :. P(k+1): 1*3+2*3^(2)+3*3^(3)+...+(k+1)*3^(k+1)=({2(k+1)-1}3^((k+1)+1)+3)/4`.
This shows that P(k+1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k+1) is true, whenever P(k) is true.
Hence, by principle of mathematical induction, we have
` 1*3+2*3^(2)+3*3^(3)+...+ n*3^(n) = ((2n-1) 3^(n+1)+3)/4 " for all values of " n in N`.
Discussion
38.

Using induction, prove that `cos theta · cos 2theta · cos 2^2 theta. ... cos 2^(n-1) theta =(sin2^n theta)/(2^n sin theta)`

Answer» Let P(n) : `costhetacos2theta. . . cos2^(n-1)theta=(sin2^(n)theta)/(2^(n)sintheta)`
Step I For n=1,P(1): `costhetacos2theta. . . cos2^(n-1)theta=(sin2^(1)theta)/(2^(n)sintheta)`
`=(sin2theta)/(2sintheta)=(2sinthetacostheta)/(2sintheta)=costheta`
which is true.
Step II Assume that P(n) is true, for n=k.
`P(k):costheta*cos2theta*cos2^(2)theta. . .cos2^(k-1)theta=(sin2^(k)theta)/(2^(k)sintheta)` is true.
Step III To prove P(k+1) is true.
`P(k=1):costheta*cos2theta*cos2^(2)theta. . .cos2^(k-1)theta*cos2^(k)theta`
`(sin2^(k)theta)/(2^(k)sintheta)*cos2^(k)theta`
`=(2sin2^(k)theta*cos2^(k)theta)/(2*2^(k)sintheta)`
`=(sin2*2^(k)theta)/(2^(k+1)sintheta)=(sin2^((k+1))theta)/(2^(k+1)sintheta)`
which is true.
So, P(k+1) is true. Hence, P (n) is true.
Discussion
39.

Prove by using the principle of mathemtical induction: ` 3.2^2 +3^2.2^3+…+3^n .2^(n+1) = 12/5 (6^n-1) `

Answer» `P(k): 3*2^(2)+3^(2)*2^(3)+…+3^(k)*2^(k+1)=12/5 (6^(k)-1)`.
Now, ` (3*2^(2)+3^(2)*3^(2)+…+3^(k)*2^(k+1))+3^(k+1)*2^(k+2)`
` ={12/5(6^(k)-1)+3^(k+1)*2^(k+1)*2}=1/5*{12(6^(k)-1)+10xx6^(k+1)}`
` = 1/5 *{2(6^(k+1)-6)+10 xx 6^(k+1)} = 12/5 (6^(k+1)-1)`.
Discussion
40.

`x^(2n-1)+y^(2n-1)` is divisible by `x+y`

Answer» ` " Let " P (n) : x^(2n)-y^(2n) " is divisible by " (x+y)`
for n=1 `P(1) : x^(2)=y^(2) =(x-y) (x+y)`
Which is divisible by x+y
`:.` P(n) is true for n=1
Let P (n) be trrue for n=1
`:. P (k) : x^(2k)-y^(2k)` ,is divisible by (x-y)
For n (K+1)
`P(K+1) : x^(2(K+1)) - y^(2(K+1))`
`=x^(2K+2)-y^(2K+2)`
`=x^(2k).x^(2)-y^(2k).y^(2)`
`=x^(2k).x^(2)-x^(2)y^(2k)+x^(2)y^2k)-y^(2k).y^(2)`
`=x^(2)(x^(2k)-y^(2k))+y^2k)(x^(2)-y^(2))`
`:. P(k)` is true
`:. x^(2)(x^(2k)-y^(2k))` is divisible by (x+y) and `y^(2k)(x-y)`
`(x+y)` is also divisible by `(x+y)`.
`:.x^(2)(x^(2k)-y^(2k))+y^(2k)(x-y)(x-y)` is divisible by (x+y)
`rArr` P (n) is also true for n=K+1
Hence from the principle of mathematical induction P (n) is true for all vlaues of n where `n in N`
Discussion
41.

Using the principle of mathematical induction. Prove that `(x^(n)-y^(n))` is divisible by (x-y) for all ` n in N`.

Answer» Let the given statement be P(n). Then,
`P(n): (x^(n)-y^(n))` is divisible by (x - y).
When n =1, the given statement becomes: `(x^(1)-y^(1))` is divisible by (x-y), which is clearly true.
` :. ` P(1) is true.
Let P(k) be true. Then,
`P(k): (x^(k)-y^(k))` is divisible by (x - y). ...(i)
Now, `(x^(k+1)-y^(y+1))`
` ={x^(k+1)-x^(k)y+x^(k)y-y^(k+1)}` [on adding and subtracting ` x^(k) y`]
` = x^(k) (x-y)+y(x^(k)-y^(k))`, which is divisible by (x - y) [using (i)].
Thus, P(k+1) is true, whenever P(k) is true.
`:. `P(1) is true and P(k+1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, it follows that
`(x^(n)-y^(n))` is divisible by (x-y) for all ` n in N`.
Discussion
42.

If `3sec^4 theta +8=10sec^2 theta` find `tan theta`.

Answer» `3sec^4theta +8 = 10sec^2theta`
`=>3sec^4theta - 10sec^2theta +8 =0`
`=>3sec^4theta - 6sec^2theta - 4sec^2theta +8 =0`
`=>3sec^2theta(sec^2theta - 2) -4(sec^2theta - 2) = 0`
`=>(3sec^2theta - 4)(sec^2theta - 2) = 0`
`=>sec^2theta = 4/3 or sec^2theta = 2`
`=>1+tan^2theta = 4/3 or 1+tan^2theta = 2`
`=>tan^2theta = 1/3 or tan^2theta = 1`
`:. tan theta = 1/sqrt3 or tan theta = 1.`
Discussion
43.

Using the principle of mathematical induction. Prove that `(1^(2)+2^(2)+…+n^(2)) gt n^(3)/3 " for all values of " n in N`.

Answer» Let `P(n): (1^(2)+2^(2)+…+n^(2)) gt n^(3)/3`.
When n =1, LHS = `1^(2) = 1 and RHS = 1^(3)/3 = 1/3`.
Since ` 1 gt 1/3`, it follows that P(1) is true.
Let P(k) be true. Then,
`P(k): (1^(2)+2^(2)+…+k^(2)) gt k^(3)/3`. ….(i)
Now, ` 1^(2)+2^(2)+...+k^(2)+(k+1)^(2) `
` ={1^(2)+2^(2)+...+k^(2)}+(k+1)^(2)`
` gt k^(3)/3 + (k+1)^(2)` [using (i)]
` = 1/3*{k^(3)+3(k+1)^(2)} = 1/3 *{k^(3)+3k^(2)+6k+3}`
` = 1/3 [{(k^(3)+1+3k(k+1)}+(3k+2)]=1/3*[(k+1)^(3) +(3k+2)]`
` gt 1/3 (k+1)^(3)`.
` :. P(k+1): 1^(2)+2^(2)+...+k^(2)+(k+1)^(2) gt 1/3 (k+1)^(3)`.
Thus, P(k+1) is true, whenever P(k) is true.
`:. ` P(1) is true and P(k+1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, it follows that
`(1^(2)+2^(2)+3^(2)+...+n^(2)) gt n^(3)/3 " for all " n in N`.
Discussion
44.

Using principle of MI prove that `2.7^n+3.5^n-5` is divisible by 24

Answer» Given expression is.
`2*7^n+3*5^n-5`
When `n = 1`, given expression is,
`2*7+3*5-5 = 12+15-5 = 24`
So, for `n=1`, given expession is divisible by `24`.
Let for any `k in N`, given expression is divisible by `24`.
Then, `2*7^k+3*5^k-5 = 24c`, where `c` is a natural number.
`=>3*5^k = 24c-2*7^k+5->(1)`
Now, we have to prove, for `n = k+1`, given expression is divisible by `24`.
For `n = k+1`, given expression is,
`2*7^(k+1)+3*5^(k+1) - 5`
`=14*7^k+5*3*5^k - 5`
From (1), given expression becomes,
`=14*7^k+5(24c-2*7^k+5) -5`
`=14*7^k+120c-10*7^k+25-5`
`=4*7^k+120c+20`
`=4*7^k+120c+24-4`
`=4(7^k -1)+24(5c+1)`
`=4((1+6)^k -1)+ 24(5c+1)`
`=4(C(k,0)6^k+C(k,1)6^(k-1)+C(k,2)6^(k-2)+...C(k,k)-1)+ 24(5c+1)
``=4(C(k,0)6^k+C(k,1)6^(k-1)+C(k,2)6^(k-2)+..+C(k,k-1)6^1.+1-1)+ 24(5c+1)
``=4(C(k,0)6^k+C(k,1)6^(k-1)+C(k,2)6^(k-2)+...6)+ 24(5c+1)
``(C(k,0)6^k+C(k,1)6^(k-1)+C(k,2)6^(k-2)+...6)` is divisible by `6`,
so we can replace this term with `6a`, where `a` is another non-zero fixed number.
`= 24a+24(5c+1)`
`=24(a+5c+1)`, which is clearly divisible by `24`.
Thus, given expression will be divisible by `24`.
Discussion
45.

Prove by the principle of induction that for all `n N , (10^(2n-1)+1)`is divisible by 11.

Answer» Let `P (n) =10^(2n-1) +1`
For n =1
`P(1)=10^(2xx1-1) +1 =10 +1 =11=11xx1`
Which is divisible by 11.
`:. `P(n) is true fon n=1
Let P (n) be true for n=K
`rArr {10^(2k-1)+1]` is divisible by 11
`:. 10^(2k-1) +1 =11 lambda , " where " lambda in I`
Now `P(k+1) =10^(2(K+1)-1)=1`
`rArr P(k+1) =10^(2k-1) . 10^(2) +1`
`=10^(2k-1).10^(2)+10^(2)-10^(2)+1`
`=10^(2)"("10^(2k-1)+1")"-100 +1`
`=100 . 11 lambda -99 =11 (100lambda-9)`
[From equation (1)]
`rArr P (K+1)` is divisible by 11
`rArr P (n) ` is also true for n=K+1
Hence by the principle of mathematical induction P(n) is true for all natural numbers n.
Discussion
46.

Prove the following by using the principle of mathematical induction for all `n in N`:`1 + 2 + 3 + ...+ n

Answer» Let the given statement be P(n) . Then,
` P(n): (1+2+3+…+n) lt 1/8 (2n+1)^(2)`.
When n =1, we have
LHS = 1 and RHS = `1/8 (2 xx 1+1)^(2) = 9/8`.
Clearly, ` 1 lt 9/8`.
`:. ` P(1) is true.
Let P(k) be true. Then,
`P(k): (1+2+3+...+k) lt 1/8 (2k+1)^(2)`. ....(i)
Now, `1+2+3+...+k+(k+1)`
` = {1+2+3+...+k} + (k+1)`
` lt 1/8 (2k+1)^(2) +(k+1) = ((2k+1)^(2)+8(k+1))/8 ` [using (i)]
` = ((4k^(2)+12k+9))/8 = ((2k+3)^(2))/8 = ({2(k+1)+1}^(2))/8 ` .
`:. {1+2+3+...+k(k+1)} lt ({2(k+1)+1}^(2))/8 `.
This shows that P(k+1) is true.
Thus, P(k+1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, it follows that
`(1+2+3+...+m) lt 1/8 (2n+1)^(2) " for all values of " n in N`.
Discussion
47.

Prove the following by using the principle of mathematical induction for all `n in N`:`1. 2+2. 2^2+3. 2^2+...+n .2^n=(n-1)2^(n+1)+2`

Answer» `P(k): 1*2+2*2^(2)+3*2^(3)+…+k*2^(k)=(k-1)*2^(k+1)+2`. …(i)
Now, `{1*2+2*2^(2)+3*2^(3)+…+k*2^(k)}+(k+1)*2^(k+1)`
` ={(k-1)*2^(k+1)+2}+(k+1)*2^(k+1)= 2k*2^(k+1)+2 = k*2^(k+1)+2`.
Discussion
48.

Prove the following by using the principle of mathematical induction for all `n in N`:`3^(2n+2)-8n-9`is divisible by 8.

Answer» Given expression is.
`3^(2n+2)-8n-9`
When `n = 1`, given expression is,
`3^4-8-9 = 81-17 = 64`
So, for `n=1`, given expession is divisible by `8`.
Let for any `k in N`, given expression is divisible by `8`.
Then, `3^(2k+2)-8k-9 = 8c->(1)`, where `c` is a natural number.
Now, we have to prove, for `n = k+1`, given expression is divisible by `8`.
For `n = k+1`, given expression is,
`3^(2(k+1)+2)-8(k+1)-9`
`= 3^2*3^(2k+2) - 8k -17`
`=9(3^(2k+2)-8k-9) +72k+81-8k-17`
`=9(3^(2k+2)-8k-9)+64k+64`
From (1),
`=9(8c)+8k(8k+8)`
`=8(9c+8k+8)`, which is clearly divisible by `8`.
Thus, given expression will be divisible by `8`.
Discussion
49.

`1^(2)+3^(2)+5^(2)+.......+(2n-1)^(2)` `=(n(2n-1)(2n+1))/(3)`

Answer» `"Let "P (n) : 1^(2)+3^(2)+5^(2)+......+(2n-1)^(2)`
` =1/3 n(4n^(2)-1)`
for n =1
`L.H.S. =1^(2) =1, R.H.S. =1/3 .1.(4.1^(2)-1)=1`
`:. L.H.S. =R.H.S.`
`rArr` P(n) is true for n=1
Let P(n) be true for n=K
`:. P(k) : 1^(2) +3^(2)+5^(2)+`
`.......+(2k-1)^(2)=1/3 k(4k^(2)-1)`
for n=K+1
`P(k+1) : 1^(2) +3^(2) +5^(2)+......+(2K-1)^(2)`
`+(2K+1)^(2)=1/3 k(4k^(2)-1)+(2k+1)^(2)`
`(k(4k^(2)-1)+3(2k+1)^(2))/(3)`
`=1/3[4k^(3)-k+12k^(2)+12k+3]`
`=1/3[4k^(3)+12k^(2)+11k+3]`
`=1/3(k+1)(4k^(2)+8k+3)`
`=1/3(k_+1)[4(k^(2)+2k+1)-1]`
`=1/3(k+1)[4(k+1)^(2)-1]`
`rArr` P (n) is also true for n=K+1
Hence from the principle of mathematical induction P (n) is true for all natural numbers. n
Discussion
50.

Prove that `2. 7^n+3. 5^n-5`is divisible by 24, for all `n in N`.

Answer» Given expression is.
`2*7^n+3*5^n-5`
When `n = 1`, given expression is,
`2*7+3*5-5 = 12+15-5 = 24`
So, for `n=1`, given expession is divisible by `24`.
Let for any `k in N`, given expression is divisible by `24`.
Then, `2*7^k+3*5^k-5 = 24c`, where `c` is a natural number.
`=>3*5^k = 24c-2*7^k+5->(1)`
Now, we have to prove, for `n = k+1`, given expression is divisible by `24`.
For `n = k+1`, given expression is,
`2*7^(k+1)+3*5^(k+1) - 5`
`=14*7^k+5*3*5^k - 5`
From (1), given expression becomes,
`=14*7^k+5(24c-2*7^k+5) -5`
`=14*7^k+120c-10*7^k+25-5`
`=4*7^k+120c+20`
`=4*7^k+120c+24-4`
`=4(7^k -1)+24(5c+1)`
`=4((1+6)^k -1)+ 24(5c+1)`
`=4(C(k,0)6^k+C(k,1)6^(k-1)+C(k,2)6^(k-2)+...C(k,k)-1)+ 24(5c+1)
``=4(C(k,0)6^k+C(k,1)6^(k-1)+C(k,2)6^(k-2)+..+C(k,k-1)6^1.+1-1)+ 24(5c+1)
``=4(C(k,0)6^k+C(k,1)6^(k-1)+C(k,2)6^(k-2)+...6)+ 24(5c+1)
``(C(k,0)6^k+C(k,1)6^(k-1)+C(k,2)6^(k-2)+...6)` is divisible by `6`,
so we can replace this term with `6a`, where `a` is another non-zero fixed number.
`= 24a+24(5c+1)`
`=24(a+5c+1)`, which is clearly divisible by `24`.
Thus, given expression will be divisible by `24`.
Discussion