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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Prove that `n(n+1)(n+5)` is a multiple of 3. |
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Answer» Let P(n) =n(n+1) (n+5) For n=1 P(1)=1.(1+1)(1+5)=12=3(4) Which is a multiple of 3. `rArr` p (n) is true for n=1 LetP (n) be true for n=K `:. P(k) =k(k+1)(K+05)=3lambda "(say) "` `" Where " lambda in I` For n=K+1 `P(k+1) =(k+1)(K=2)(K+6)` `=(K+1)[K^(2)+8K+12]` `=(k=1)[K^(2)=5k)+(3k+12)]` `=(k+1)k(k+5)+3(k+1)(+4)` `=3lamda+3(K+1)(K+4)` [From equation (1)] `=3[lambda+(k=1)(K+4)]` Which is a multiple of 3. `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. |
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| 52. |
`(1+(1)/1)(1+(1)/(2))(1+(1)/(3))......(1+(1)/n)` `n(n+1)` |
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Answer» Let `P(n): (1+1) (1+(1)/(2))(1+(1)/(3))......(1+(1)/(n)) =(n+1)` for n=1 `L.H.S. =1+1=2` `R.H.S. =1+1=2` `:. 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+1)(1+(1)/(2))(1+(1)/(3))......(1+(1)/(k)) =K+1` for n = K+1 `P(k+1) : (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` `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 . |
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| 53. |
Prove that `1+2+3+4........+N |
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Answer» `" Let " P(n) : 1+2+3 +....+ nlt1/8(2n+1)^(2)` for n=1 `L.H.S. =1, R.H.S. =1/8 (2+1)^(2)=9/8` `:. L.H.S. lt R.H.S. ` `rArr` P(n) is true for n=1 Let P(n) be true for n=K `P(k) : 1 +2+3+….+klt 1/8 (2k+1)^(2) ……(1)` `n=K+1` `P(k+1) : 1+2+3 +.....+K+(K+1)` ` lt 1/8 (2k+1)^(2)+(K+1) " "["From eauation "(1)]` `=((2k+1)^(2)+8(k+1))/(8)` `=(4k^(2)+4k+1+8K+8)/(8) =(4k^(2)+12K+9)/(8)` `=1/8 (2k+2)^(2) =1/8 {2(k+1)+1}^(2)` `rArr` P(n) is also true for n=K+1 Hence from the principle of mathematical induction the given statement is true for all `n in N` |
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| 54. |
Prove that `1^2+2^2+dotdotdot+n^2>(n^3)/3,``n in N` |
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Answer» We will prove it using mathematical induction. For, `n = 1`, `L.H.S = 1^2 = 1` `R.H.S. = 1^3/3 = 1/3` As, `1 gt 1/3`, our equation is true for `n = 1`. Let, our equation is true for `n = k` where `k` is a natural number. Then, `1^2+2^2+3^2+...k^2 gt k^3/3->(1)` Now, we have to prove, for `n = k+1`, given equation is true. For, `n = k+1`, `L.H.S. = 1^2+2^2+3^2+...k^2+(k+1)^2` From (1), `1^2+2^2+3^2+...k^2+(k+1)^2 gt k^3/3+(k+1)^2` `gt 1/3(k^3+3k^2+6k+3)` `gt 1/3((k^3+1^3+3k(k+1))+(3k+2))` `gt 1/3((k+1)^3)+(3k+2))` `gt 1/3(k+1)^3+1/3(3k+2)` `gt 1/3(k+1)^3`, as `1/3(3k+2) gt 0` `:. 1^2+2^2+3^2+...k^2+(k+1)^2 gt 1/3((k+1)^3)` Thus, our equation is true for `n = k+1`. `:. 1^2+2^2+3^2+...n^2 gt n^3/3` |
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| 55. |
Prove the following by using the principle of mathematical induction for all `n in N`:`(1+3/1)(1+5/4)(1+7/9)dotdotdot(1+((2n+1))/(n^2))=(n+1)^2` |
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Answer» `" Let "P(n) : (1+3)(1+(5)/(4))(1+(7)/(9))` `......(1+(2n+1)/(n^(2)))=(n+1)^(2)` For n=1 `L.H.S. =1+3 =4,R.H.S. =(1+1)^(2)=4` `:. 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=3)(1+(5)/(4)) (1+(7)/(9))` `......(1+(2k+1)/(k^(2))) =(k+1)^(2)` For n=K+1 `P(k+1) : (1+3) (1+(5)/(4)) (1+(5)/(9))` `......(1+(2K+1)/(k^(2))).{1+(2(k+1)+1)/((k+1))}` `=(k+1)^(2) {1+(2(K+1)+1)/((K+1)^(2))}` `=(k+1)^(2){((k+1)^(2)+2(K+1)+1)/((k+1)^(2)]}` `={(K+1)+1}^(2)=(K+2)^(2)` `rArr` P (n) is also frue for n=K+1 Hence from the principle of the mathamatical induction P (n) is true for all natural numbers n. |
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| 56. |
Using the principle of mathematical induction, prove that `(1-1/2)(1-1/3)(1-1/4)...(1-1/(n+1))= 1/((n+1))" for all " n in N`. |
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Answer» Let the given statement be P(n). Then, `P(n) : (1-1/2)(1-1/3)(1-1/4)...(1-1/(n+1))= 1/((n+1))` When ` n = 1, LHS = (1-1/2) = 1/2 and RHS = 1/((1+1)) = 1/2`. ` :. ` LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, ` P(k): (1-1/2)(1-1/3)(1-1/4)...(1-1/(k+1)) = 1/((k+1))`. ...(i) Now, `{(1-1/2)(1-1/3)(1-1/4)...(1-1/(k+1))}*(1-1/(k+2))` ` = 1/((k+1))*[((k+2)-1)/((k+2))]=1/((k+1))*((k+1))/((k+2)) = 1/((k+2))` [using (i)]. ` :. P(k+1):(1-1/2)(1-1/3)(1-1/3)(1-1/4)...(1-1/(k+2))= 1/((k+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 `(1-1/2)(1-1/3)(1-1/4)...(1-1/(n+1)) = 1/((n+1)) " for all " n in N`. |
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| 57. |
Prove the following by using the principle of mathematical induction for all `n in N`:`1^2+3^2+5^2+dotdotdot+(2n-1)^2=(n(2n-1)(2n+1))/3` |
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Answer» `P(k): 1^(2)+3^(2)+5^(2)+…+(2k-1)^(2)=(k(2k-1)(2k+1))/3 `. Now, `{1^(2)+3^(2)+5^(2)+…+(2k-1)^(2)}+{2(k+1)-1}^(2)` `=(k(2k-1)(2k+1))/3 +(2k+1)^(2)= 1/3 *{k(2k-1)(2k+1)+3(2k+1)^(2)}` ` = 1/3 (2k+1){k(2k-1)+3(2k+1)}=1/3 (2k+1)(2k^(2)+5k+3)` ` = 1/3 (k+1)(2k+1)(2k+3)`. |
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| 58. |
Using the principle ofmathematical induction prove that`1/(1. 2. 3)+1/(2. 3. 4)+1/(3. 4. 5)++1/(n(n+1)(n+2))=(n(n+3))/(4(n+1)(n+2)`for all `n in N` |
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Answer» Let the given statement be P(n). Then, `P(n): 1/(1*2*3)+1/(2*3*4)+…+1/(n(n+1)(n+2)) = (n(n+3))/(4(n+1)(n+2)).` Putting n = 1 in the given statement , we get LHS = ` 1/(1*2*3) = 1/6 and RHS = (1xx(1+3))/(4xx(1+1)(1+2)) = (1xx4)/(4xx2xx3) = 1/6 `. `:." " ` LHS=RHS. Thus, the given statement is true for n = 1, i.e., P(1) is true. Let P(k) be true. Then, `P(k): 1/(1*2*3)+1/(2*3*4)+...+1/(k(k+1)(k+2))=(k(k+3))/(4(k+1)(k+2)).`...(i) Now, `1/(1*2*3)+1/(2*3*4)+...+1/(k(k+1)(k+2))+1/((k+1)(k+2)(k+3)) ` `={1/(1*2*3)+1/(2*3*4)+...+1/(k(k+1)(k=2))}+1/((k+1)(k+2)(k+3)) ` `={(k(k+3))/(4(k+1)(k+2))+1/((k+1)(k+2)(k+3))}" "`[using (i)] `=(k(k+3)^(2)+4)/(4(k+1)(k+2)(k+3))=((k^(3)+6k^(2)+9k+4))/(4(k+1)(k+2)(k+3)) ` ` = ((k+1)(k+1)(k+4))/(4(k+1)(k+2)(k+3))= ((k+1)(k+4))/(4(k+2)(k=3))`. `:." " P(k+1): 1/(1*2*3)+1/(2*3*4)+...+1/((k+1)(k+2)(k+3))=((k+1)(k+4))/(4(k+2)(k+3))`. 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/(1*2*3)+1/(2*3*4)+1/(3*4*5)+...+1/(n(n+1)(n+2)) = (n(n+3))/(4(n+1)(n+2)) " for all values of " n in N`. |
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| 59. |
Prove the following by using the principle of mathematical induction for all `n in N`:`a+a r+a r^2+dotdotdot+a r^(n-1)=(a(r^n-1))/(r-1)` |
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Answer» Let the given statement be P(n) . Then , `P(n): a + ar + ar^(2) + …+ar^(n-1)= (a(r^(n)-1))/((r-1))`. When n = 1, we have LHS = a and RHS = `(a(r^(1)-1))/((r-1))`. When n = 1, we have LHS = a and RHS = `(a(r^(1)-1))/((r-1)) = a`. `:. `LHS = RHS. Thus, P(1) is true. LetP(k) be true. Then, `P(k) : a+ar+ar^(2)+...+ar^(k-1)=(a(r^(k)-1))/((r-1))`. ...(i) Now, `(a+ar+ar^(2)+...ar^(k-1)) + ar^(k) = (a(r^(k)-1))/((r-1)) + ar^(k)` [using (i)] `=(a(r^(k+1)-1))/((r-1))`. `:. P(k+1): a+ar+ar^(2)+...+ar^(k-1)+ar^(k) = (a(r^(k+1)-1))/((r-1))`. 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 ` a+ar+ar^(2) +...+ ar^(n-1) = (a(r^(n)-1))/((r-1)) " for " r gt 1 and " all "n in N`. |
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| 60. |
Prove the following by the principle ofmathematical induction: `n^3-7n+3`is divisible 3 for all `n in N`. |
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Answer» Let P(n) : `n^(3)-7n+3` is divisible by 3, for all natural number n. Step I We observe that P(1) is true. `P(1)=(1)^(3)-7(1)+3` =11-7+3 -3, which is divisible by 3. Hence, P(1) is true. Step II Now, assume that P(n) is true for n=k. `P(k+1):(k+1)^(3)-7(k+1)+3` `=k^(3)+1+3k(k+1)-7k-7+3` `=k^(3)-7k+3+3k(k+1)-6` `=3q+3[k(k+1)-2]` Hence, P(k+1) is true whenever P(k) is true. [from stem II] So, by the principle of mathematical induction P(n): is true for all natural number n. |
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| 61. |
Prove the following by the principle ofmathematical induction: `1/(3. 5)+1/(5. 7)+1/(7. 9)+1/((2n+1)(2n+3))=n/(3(2n+3))` |
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Answer» Let the given statement be P(n). Then, ` P(n) : 1/(3*5)+1/(5*7) + 1/(7*9) +…+ 1/((2n+1)(2n+3)) = n/(3(2n+3))`. Putting n = 1 in the given statement, we get LHS ` = 1/(3*5) = 1/15 and RHS = 1/(3(2xx1+3))= 1/15`. ` :. ` LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, `P(k): 1/(3*5)+1/(5*7)+1/(7*9) + ...+1/((2k+1)(2k+3)) = k/(3(2k+3)).`...(i) Now, ` 1/(3*5) + 1/(5*7) + ...+1/((2k+1)(2k+3))+1/({2(k+1)+1}{2(k+1)+3}) ` `{1/(3*5)+1/(5*7)+..+1/((2k+1)(2k+3))}+1/((2k+3)(2k+5))` ` = k/(3(2k+3))+1/((2k+3)(2k+5)) " "` [using (i)] `=(k(2k=5)+3)/(3(2k+3)(2k+5))= ((2k^(2)+5k+3))/(3(2k+3)(2k+5))=((k+1)(2k+3))/(3(2k+3)(2k+5)) ` `=((k+1))/(3(2k+5))= ((k+1))/(3{2(k+1)+3}).` `:. " " P(k+1) : 1/(3*5)+1/(5*7)+1/(7*9)+...+1/({2(k+1)+1}{2(k+1)+3}) = ((k+1))/(2{2(k+1)+3}).` 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*5)+1/(5*7)+1/(7*9)+...+ 1/((2n+1)(2n+3)) = n/(3(2n+3)) " for all values of " n in N`. |
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| 62. |
Prove the following by the principle ofmathematical induction:` (a b)^n=a^n b^n`for all `n in N`. |
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Answer» Let the given statement be P(n). Then, `P(n): (ab)^(n) = a^(n) b^(n)`. When n = 1 , we have LHS = `(ab)^(1) = ab and RHS = a^(1)b^(1) = ab`. `:. ` LHS = RHS. Thus , the given statement is true for n = 1, i.e., P(1) is true. Let P(k) be true. Then, `P(k) : (ab)^(k) = a^(k) b^(k)`. ....(i) Now, `(ab)^(k+1) = (ab) ^(k) (ab) = (a^(k)b^(k))(ab)` [using (i)] ` = (a^(k)*a)(b^(k)*b)` [by commnutativity and associativity of multiplication on real numbers] `(a^(k+1)*b^(k+1))`. `:. " " P(k+1): (ab)^(k+1)=(a^(k+1)*b^(k+1))`. 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 `(ab)^(n) = a^(n) b^(n) " for all " x in N`. |
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| 63. |
`1.3+2.3^(2)+3.3^(3)+. . .+n.3^(3)=((2n-1)3^(n+1)+3)/(4)` |
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Answer» `P(n) : 1 xx 3 xx + 2 xx 3^(2) + 3 xx 3^(3) + "…." n xx 3^(n)` `= ((2n-1) 3^(n+1) + 3)/(4)` For `n = 1` and `L.H.S. = 1 xx 3 = 3` and `R.H.S. = ((2 xx 1 - 1)3^(1+1)+3)/(4) = (3^(2)+3)/(4) = (12)/(4) = 3` Thus `P(1)` is true. Let `P(n)` be true for some `n = k` i.e, `1 xx 3 + 2 xx 3^(2) + 3 xx 3^(3) + "..... + k xx 3^(k)` `= ((2k-1)3^(k+1)+3)/(4)` Now, we have to prove that `P(n)` is true for `n = k 1` ltb rgt i.e, `1 xx 3 + 2 xx 3^(2) + 3 xx 3^(3) + "...." + k +1 xx 3^(k+1)` `= ((2k+1)3^(k+2)+3)/(4)` Adding `(k+1) xx 3^(k+1)` both sides of `(1)`, we get `1 xx 3 + 2 xx 3^(2) + 3 xx 3^(3) + "...." + k xx 3^(4) + (k+1) xx 3^(k+1)` `= ((2k-1)3^(k+1)+3)/(4)+(k+1)xx3^(k+1)` `= ((2k-1)3^(k+1)+3+4(k+1)3^(k+1))/(4)` `= (3^(k+1)[2k-1+4(k+1)]+3)/(4)` `= (3^(k+1)(6k+3)+3)/(4)` ` = (3^((k+1)+1)(2k+1)+3)/(4)` `= ((2k+1)3^(k+2)+3)/(4)` Thus `P (k+1)` is true true whenever `P(k)` is true. Hence by the principle of mathematical induction, statement `P(n)` is true for all natural numbers. |
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| 64. |
Using the principle of mathematical induction, prove that ` 1/(1*2)+1/(2*3)+1/(3*4)+…+1/(n(n+1)) = n/((n+1)) `. |
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Answer» Let the given statement be P(n). Then, ` P(n): 1/(1*2)+1/(2*3)+1/(3*4)+..+1/(n(n+1))= n/((n+1))`. Putting n = 1 in the given statement, we get LHS ` = 1/(1*2) = 1/2 and RHS = 1/((1+1)) = 1/2`. `:. ` LHS= RHS. Thus, P(1) is true. Let P(k) be true. Then, `P(k) : 1/(1*2)+1/(2*3)+1/(3*4) +...+ 1/(k(k+1)) = k/((k+1)) `. ...(i) Now, ` 1/(1*2)+1/(2*3)+1/(3*4)+...+1/(k(k+1)) + 1/((k+1)(k+2)) ` ` = {1/(1*2)+1/(2*3)+1/(3*4)+...+1/(k(k+1))} + 1/((k+1)(k+2)) ` ` = k/((k+1)) + 1/((k+1)(k+2))" "` [using (i)] ` =(k(k+2)+1)/((k+1)(k+2)) = ((k+1)^(2))/((k+1)(k+2)) = ((k+1))/((k+2))`. `:. " " P(k+1) : 1/(1*2)+1/(2*3)+1/(3*4)= 1/((k+1)(k+2)) = ((k+1))/((k+2)) .` 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, we have ` 1/(1*2)+1/(2*3)+1/(3*4) +...+ 1/(n(n+1)) = n/(n+1)" for all " n in N`. |
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| 65. |
Using the principle of mathematical induction, prove that `(n^(2)+n)` is seven for all ` n in N`. |
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Answer» Let P(n) : `(n^(2)+n)` is even. For n=1, the given expression becomes `(1^(2)+1) = 2`, which is even. So, the given statement is true for n = 1, i.e., P(1) is true. Let P(k) be true. Then, `P(k):(k^(2)+k)` is even ` rArr" " (k^(2)+k) = 2m` for some natural number m. ....(i) Now, ` (k+1)^(2)+(k+1)=k^(2)+3k+2=(k^(2)+k)+2(k+1)` ` = 2m + 2(k+1)" "` [using (i) ] ` =2[m+(k+1)]`, which is clearly even. ` :. P(k+1):(k+1)^(2)+(k+1)` is even. 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 induction, it follows that `(n^(2)+n)` is even for all ` n in N`. |
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| 66. |
Using the principle of mathematical induction, prove that ` (7^(n)-3^(n))` is divisible by 4 for all ` n in N`. |
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Answer» Let `P9n): (7^(n)-3^(n))` is divisible by 4. For n = 1, the given expression becomes `(7^(1)-3^(1))= 4`, which is divisible by 4. So, the given statement is true for n = 1 , i.e., P(1) is true. Let P(k) be true. Then, `P(k): (7^(k)-3^(k))` is divisible by 4. ` rArr (7^(k)-3^(k)) = 4m` for some natural number m. ....(i) Now, `{7^((k+1))-3^((k+1))}` `=7^((k+1))-7*3^(k)+7*3^(k)-3^((k+1))" "` [subtracting and adding `7*3^(k)`] ` = 7(7^(k)-3^(k))+3^(k)(7-3)` ` =(7xx4m)+4*3^(k)` [using (i)] ` = 4(7m+3^(k))`, which is clearly divisible by 4. ` :. P(k+1):{7^((k+1))-3^((k+1))}` is divisible by 4. 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 `(7^(n)-3^(n))` is divisible by 4 for all values of ` n in N`. |
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| 67. |
Prove the following by using the principle of mathematical induction for all `n in N`:`n(n + 1) (n + 5)`is a multiple of 3. |
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Answer» Let P(n): n(n+1)(n+5) is a multiple of 3. For n = 1, the given expression becomes `(1xx2xx6) = 12`, which is a multiple of 3. So, the given statement is true for n = 1, i.e., P(1) is true. Let P(k) be true. Then, `P(k) : k(k+5)(k+5)` is a multiple of 3 ` rArr k(k+1)(k+5) = 3m` for some natural number m. ...(i) Now, `(k+1)(k+2)(k+6)` ` =(k+1)(k+2)k+6(k+1)(k+2)` `=k(k+1)(k+5-3)+6(k+1)(k+2)` ` = k(k+1)(k+5)-3k(k+1)+6(k+1)(k+2)` ` = 3m+3(k+1)(2k+4-k)` ` = 3m+3(k+1)(k+4)` ` =3{m+(k+1)(k+4)}`, which is a multiple of 3. ` :. P(k+1):(k+1)(k+2)(k+6)` is a multiple of 3. Thus, P(k+1) is true, whenever P(k) is true. `:. ` P(1) is true and P(k+1) is ture, whenever P(k) is true. Hence , by the principle of mathematical induction, it follows that n(n+1)(n+5) is a multiple of 3 for all ` n in N`. |
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| 68. |
Prove the following by using the principle of mathematical induction for all `n in N`:`1. 2 + 2. 3 + 3. 4 + ... + n(n + 1)=[(n(n+1)(n+2))/3]` |
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Answer» Let the given statement be P(n). Then, `P(n): 1*2+2*3+3*4+..+n(n+1)= 1/3 n(n+1)(n+2)`. When n = 1, LHS = ` 1*2=2 and RHS = 1/3 xx 1 xx 2 xx(1+2)=2`. `:. ` LHS = RHS. Thus, the given statement is true for n = 1, i.e., P(1) is true. Let P(k) be true. Then, `P(k) : 1*2+2*3+3*4+...+k(k+1) = 1/3 k(k+1)(k+2)`. ....(i) Now, `1*2+2*3+3*4+...+k(k+1)+(k+1)(k+2)` ` = {1*2+2*3+3*4+...+k(k+1)+(k+1)(k+2)` ` = 1/3 k(k+1)(k+2)+(k+1)(k+2)" "` [using (i)] ` = 1/3 *[k(k+1)(k+2)+3(k+1)(k+2)]= 1/3 (k+1)(k+2)(k+3)`. ` :. " " P(k+1) : 1* 2 + 2 + 2 * 3 + 3 * 4 +...+(k+1)(k+2)=1/3 (k+1)(k+2)(k+3)`. 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*2+2*3+3*4+...+n(n+1) = 1/3 n(n+1) (n+2)" for all " n in N`. |
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| 69. |
Prove the following by using the principle of mathematical induction for all `n in N`:`41^n-14^n`is a multiple of 27. |
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Answer» Let ` ((41)^(k)-(14)^(k))/27 = p." "`…(i). Then, `{(41)^(k+1)-(14)^(k+1)}= (41)^(k+1)-(41)^(k)*14+(41)^(k)*14-(14)^(k+1)` ` = (41)^(k)(41-14)+14{(41)^(k)-(14)^(k)} = 27 xx (41)^(k)+14xx 27p` ` = 27 xx [(41)^(k) + 14]`. |
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| 70. |
Using the principle of mathematical induction , prove that for `n in N`, `41^(n) - 14^(n)` is a multiple of 27. |
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Answer» `P(n) : 41^(n)-14^(n) = 27x, x in N` For `n = 1 , 41^(1) - 14^(1) = 27` So, `P(1)` is true, Let `P(k)` be true. i.e, `41^(k) - 14^(k) = 27m, m in N"….."(1)`, Now, we have to prove that `P(k+1)` is true. i.e, `41^(k+1)-14^(k+1)=27p, p in N` `41^(k+1)-14^(k+1)` `= 41 xx 41^(k) - 14 xx 14^(k)` ` = 41 xx (41^(k) - 14^(k) + 14^(k) xx 14)` `= 41(41^(k) - 14^(k)) + 41 xx 14^(k) - 14^(k) xx 14` ` = 41 xx 27 m + 27 xx 14^(k)` [Using `(1)`] `= 27 xx (41m + 14^(k))` `= 27 xx q [ q = (41m + 14^(k))]` thus, `P(k+1)` is true whenever `P(k)` is true. Hence, by the principle of mathematical induction, statement `P(n)` is true for all natural numbers. |
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| 71. |
Using principle of mathematical induction, prove that for all `n in N, n(n+1)(n+5)` is a multiple of 3. |
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Answer» We have: ` P(n) : n(n+1)(n+5) = 3x, x in N` For `n = 1, 1 xx (1+1)(1+5) = 12`, which is a multiple of 3. So, `P(1)` is true. Let `P(n)` be true for some `n = k`. i.e., `k(k+1) (k+5) = 3m`, where `m in N"….."(1)` We have to prove that `P(k+1)` is true. i.e., `(k+1)(k+2)(k+6)=3p, p in N` Now, `(k+1)(k+2)(k+6)-k(k+1)(k+5)` `= (k+1)[(k+2)(k+6)-k(k+5)]` `= (k+1)[k^(2)+8k+12-k^(2)-5k]` ltbr gt `= (k+1)(3k+12)` `= 3q`, where `q in N` Thus, `(k+1)(k+2)(k+6)` `= k (k+1)(k+5)+3q` `= 3m + 3q`, which is multiple of 3. Thus, `P(k+1)` is true whenever `P(k)` is true. Hence, by the principle of mathematical induction, statement `P(n)` is true for all natural numbers. |
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| 72. |
Prove the following by the principle ofmathematical induction: ` 1. 3+2. 4+3. 5++(2n-1)(2n+1)=(n(4n^2+6n-1))/3` |
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Answer» Let the given statement be P(n). Then, `P(n): 1*3+3*5+5*7+…+(2n-1)(2n+1)= 1/3 n(4n^(2)+6n-1)`. When n = 1, we have LHS = ` 1*3 and RHS = 1/3 xx 1 xx(4xx1^(2)+6xx1-1)=1/3 xx 1 xx 9 = 3`. ` :. ` LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, `P(k): 1*3+3*5+5*7+...+(2K-1)(2k+1)= 1/3 k(4k^(2)+6k-1)." "` ...(i) Now, ` 1*3+3*5+5*7+...+(2k-1)(2k+1)+{2(k+1)-1}{2(k+1)+1}` ` ={1*3+3*5+5*7+...+(2k-1)(2k+1)}+(2k+1)(2k+3)` ` = 1/3 k(4k^(2)+6k-1)+(2k+1)(2k+3)" "`[using (i)] ` = 1/3 [(4k^(3)+6k^(2)-k)+3(4k^(2)+8k+3)] = 1/3 (4k^(3)+18k^(2)+23k+9)` ` = 1/3 (k+1)(4k^(2)+14k+9)= 1/3 (k+1)[4(k+1)^(2)+6(k+1)-1]`. ` :. P(k+1): 1*3+3*5+5*7+...+{2(k+1)-1}{2(k+1)+1}` ` = 1/3 (k+1){4(k+1)^(2)+6(k+1)-1}`. 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 , we have ` 1*3+3*5+5*7+...+(2n-1)(2n+1) = 1/3 n (4n^(2)+6n-1)" for all " n in N`. |
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| 73. |
Prove by the principle of mathematical induction that for all `n in N :``1^2+2^2+3^2++n^2=1/6n(n+1)(2n+1)` |
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Answer» Let the given statement be P(n). Then, `P(n): 1^(2)+2^(2)+3^(2)+…+n^(2)= 1/6 n(n+1)(2n+1)`. Putting n = 1 in the given statement, we get LHS = ` 1^(2) = 1 and RHS = 1/6 xx 1 xx 2(2xx 1+1) = 1`. ` :." " `LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, `P(k): 1^(2)+2^(2)+3^(2)+...+k^(2)= 1/6 k(k+1)(2k+1)`. ...(i) Now, `1^(2)+2^(2)+3^(2)+...+k^(2)+(k+1)^(2)` ` = {1^(2)+2^(2)+3^(2)+...+k^(2)}+(k+1)^(2)` ` = 1/6 k(k+1)(2k+1)+(k=1)^(2)" "` [using (i)] `=1/6 (k+1)*{k(2k+1)+6(k+1)}` ` = 1/6 (k+1)(2k^(2)+7k+6) = 1/6 (k+1){(2k^(2)+4k)+(3k+6)}` ` = 1/6 (k+1)(k+2)(2k+3)= 1/6 (k+1)(k+1)[2(k+1)+1]`. ` :." " P(k+1):1^(2)+2^(2)+3^(2)+...+(k+1)^(2)` ` " "= 1/6 (k+1)(k+1+1)[2(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^(2)+2^(2)+3^(2)+...+n^(2) = 1/6 n(n+1) (2n+1)" for all " n in N`. |
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| 74. |
`1/2+1/4+1/8+1/16+.......+1/2^n=1-1/2^n` |
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Answer» `" Let "P(n) : 1/2+1/4+1/8+.........+1/2^(n)=1-1/2^(n)` for n=1 `L.H.S. =1/2` `R.H.S. =1-1/2^(1)` `=1-1/2=1/2` `:. " "L.H.S. =R.H.S.` therefore P (n) is true for n=1. Let P (n) true for n=k `P (n) : 1/2+1/4+1/8+......+1/2^(k) =1-1/2^(k)` for n=K+1 `P(k+1) : 1/2+1/4+1/8+......+1/2^(k)+1/2^(k+1) =1-1/2^(k)+1/2^(k+1)` `=1-((1)/(2^(k))-(1)/(2^(k+1)))` `=1-((2-1)/(2^(k+1)))=1-(1)/(2^(K+1))` `rArr` P (n) is also true for n=K+1 Hence from the principle of mathematical induction the given statement is true for all natural numbers n. |
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| 75. |
`1. 2 .3+2. 3 .4++n(n+1)(n+2)=(n(n+1)(n+2)(n+3))/4` |
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Answer» Let P (n) `: 1.2.3+2.3.4+3.4.5+……` `+n(n+1)(n+2)=1/4n(n+1)(n+2)(n+3)` For n=1, `L.H.S. =1.2.3=6` `R.H.S. =1/4 .1.(1+1)(1+2)(1+3)=6` `:. 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 + 2.3.4+3.4.5+......` `.......+k(k=1)(k+2)` `=1/4 k(k+1)(k+2)(k+3)` For n=k+1 `P (k+1) : 1.2.3+2.3.4+3.4.5+........+k(k+1)` `(k+2)+(k+1)(k+2)(K+3)` `=1/4k(k+1)(k+2)(K+3)+(K+1)(K+2)(K+3)` `=(k+1)(K+2)(K+3) ((K)/(4)+1)` `=1/4 (k+1) (k+2) (k+3) (k+4)` `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. |
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| 76. |
`1.3+2.3^2+3.3^3+..............+n.3^n=((2n-1)3^(n+1)+3)/4` |
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Answer» For n=1 `L.H.S.=1.3=3` , `R.H.S. =((2-1).3^(2)+3)/(4)` `=(9+3)/(4)=3` ` :. L.H.S. =R.H.S.` Therefore given statements is true for n=1 Let the statement be true for n=k. `:. 1.3+2.3^(2)+3.3^(3)+…….+k.3^(k)` `=((2k -1)3^(k+1)+3)/(4)` For n =k +1 `1.3+2.3^(2)+3.3^(3)+.....+k.3^(k)+(k+1).3^(k+1)` `=((2k-1)3^(k+1)+3)/(4) +(k+1).3^(k+1)` [From equation (a)] `=((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)` `=(3(2k+1).3^(k+1)+3)/(4)` `=((2K+1).3^(k+2)+3)/(4)` `rArr` statement is also true for n=k+1 Hence form the principle of mathematical induction the given statement is true for all natural numbers n |
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| 77. |
Prove by PMI that `1.2+ 2.3+3.4+....+ n(n+1) =((n)(n+1)(n+2))/3, AA n in N` |
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Answer» Let `p (n) : 1.2 +2.3+3.4+…..+n.(n+1)` `=1/3n(n+1)(n+2)` For n =1 `L.H.S. =1.2=2` `R.H.S. =1/3.1.(1+1)(1+2)=2` `:. 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+2.3+3.4+....+k.(k+1)` `=1/3 k(k +1)(k+2)` For n=K+1 `P(k +1) :1.2+2.3+3.4+....+k.(k+1)+(k+1)(K+2)` `=1/3 k(k+1) (K+2)+(K+1)(K+2)` `=(k+1)(k+2) ((1)/(3)k+1)` `=((k+1)(K+2)(k+3))/(3)` `rArr` P (n) is also true for n=(K+1) Hence from the principle of mathematical indicution P (n) is true for all natural numbers n. |
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| 78. |
Using the principle ofmathematical induction, prove that `(2^(3n)-1)`is divisible by `7`for all `n in Ndot` |
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Answer» Let `P(n) : 2^(3n) - 1` is divisible by 7. For `n = 1, 2^(3) - 1 = 7` , which is divisible by `7`. Thus `P(1)` is true. Let `P(n)` be true for some `n =k`. Then `2^(3k) - 1 = 7m, m in N"………."(1)` Now, `2^(3(k+1)) - 1 = 8 xx 2^(3k) - 1` `= 8(7m+1)-1` [Using `(1)`] `= 56m + 7` `= 7(8m+1)`, which is divisible by 7. Thus `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. |
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