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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Let `I_(m)=int_(0)^(pi)((1-cos mx)/(1-cos x))dx` use mathematical induction to prove that `l_(m)= m pi, m=0,1,2`...... |
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Answer» `becauseI_(m)=int_(0)^(pi)((1-cosmx)/(1-cosx))dx` Step I For `m=1,l_(1) int_(0)^(pi)((1-cosx)/(1-cosx))dx` `therefore I_(1) = pi and "for" m=2`, `I_(2)int_(0)^(pi)((1-cos2x)/(1-cosx))dx` `=int_(0)^(pi)(2sin^2x(1+cosx))/((1-cosx)(1+cosx))dx` `=int_(0)^(pi)(2sin^2x(1+cosx))/(sin^2x)dx=2int_(0)^(pi)(1+cosx)dx` `=2[x + sin x]_(0)^(pi)=2(pi+0)-(0+0)=2pi` which are true , therefore , `I_1 and I_2` are true. Step II Assume `I_(k+1)=int_(0)^(pi)(1-cos(k+1)x)/(1-cosx)dx` `therefore I_(k+1)-I_(k)=int_(0)^(pi)(coskx-cos (k+1)x)/(1-cosx)dx` `=int_(0)^(pi)(2sin((2k+1)/(2))x.sin((x)/(2)))/(2sin^2((x)/(2)))dx` `=int_(0)^(pi)(sin((2k+1)/2)x)/(sin((x)/(2)))dx`.....(iii) Similarly ,`I_(k)-I_(k-1)=int_(0)^(pi)(sin((2k-1)/(2))x)/(sin((x)/(2)))dx`....(iv) On subtracting Eq.(iv) from Eq.(iii) , we get `I_(k+1)` -2I_(k)+I_(k-1)=int_(0)^(pi)(sin((2k+1)/(2))x-sin((2k-1)/(2))x)/(sin((x)/(2)))dx` `=int_(0)^(pi)(2cos(kx)sin((x)/(2)))/(sin((x)/(2)))dx2 int_(0)^(pi)cos kxdx =2[(sinkx)/(k)]_(0)^(pi)=0` `rArr l_(k+1)=2I_(k)-I_(k-1)=2kpi-(k-1)pi` [by assumption step] `=kpi+pi=(k=1)pi` This show that the result is true for `m=k+1`. Hence , by the principle of mathematical induction the result is true for all `m in N`. |
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| 2. |
Prove by mathematical induction that `(1)/(1+x)+(2)/(1+x^2)+(4)/(1+x^4)+.....+(2^n)/(1+x^(2^n))=(1)/(x-1)+(2^(n+1))/(1-x^(2^(n+1)))` where , `|x|ne 1 ` and n is non - negative integer. |
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Answer» Let`P(n):(1)/(1+x)+(2)/(1+x^2)+(4)/(1+x^4)+.....+(2^n)/(1+x^(2^(n)))` `=(1)/(x-1)+(2^(n+1))/(1-x^(2^(n+1)))` .....(i) Step I For `n=1`, LHS of Eq. (i) =(1)/(1+x)+(2)/(1+x^2)` |
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| 3. |
Using the principle of mathematical induction to prove that `int_(0)^(pi//2)(sin^2nx)/(sinx)dx=1+(1)/(3)+(1)/(5)+.....+(1)/(2n-1)` |
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Answer» Let `P(n): int _(0)^(pi//2)(sin^2nx)/(sinx)dx=1+(1)/(3)+(1)/(5)+......+(1)/(2n-1)` ..........(i) Step I For n =1 LHS of Eq. 9i) `= int_(0)^(pi//2)(sin^2x)/(sinx)dx=int_(0)^(pi//2)sin xdx=-[cosx]_(0)^(pi//2)=-(0-1)=1` and RHS of Eq. (i) =1 Therefore , P(1) is true . Step II Assume it is true for n=k, then `P(k):int_(0)^(pi//2)(sin^2kx)/(sinx)dx=1+(1)/(3)+(1)/(5)+.......+(1)/(2k-1)` Step III For `n=k+1`, `P(k+1):int_(0)^(pi//2)(sin^2(k+1)x)/(sinx)dx=1+(1)/(3)+(1)/(5)+.....+(1)/(2k-1)+(1)/(2k+1)` LHS `=int_(0)^(pi//2)(sin(k+1)x)/(sinx)kdx` `=int_(0)^(pi//2)(sin^2(k+1)x-sin^2kx+sin^2kx)/(sinx)dx` `=int_(0)^(pi//2)(sin^2(k+1)x-sin^2kx)/(sinx)dx+int_(0)^(pi//2)(sin^2kx)/(sinx)dx` `=int_(0)^(pi//2)(sin(2k+1)xsinx)/(sinx)dx+P(k)` [by assumption step] `=int_(0)^(pi//2)sin(2k+1)xdx+P(k)` `=-[(cos (2k+1)x)/(2k+1)]_(0)^(pi//2)+P(k)` `=-(1)/((2k+1))[cos(pik+(pi)/(2))-1]+P(k)` `=-(1)/((2k+1))[-sinpik-1]+P(k)` `=-(1)/(2k+1)[-0-1]+P(k)` `=(1)/((2k+1))+1+(1)/(3)+(1)/(5)+......+(1)/((2k-1))` [by assumption step] `=1+(1)/(3)+(1)/(5)+.....+(1)/((2k-1))+(1)/((2k+1))=RHS` This shows that the result is true for `n=k+1`. Hence , by the principle of mathematical induction , the result is true for all `n in N`, |
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| 4. |
`n^7-n` is divisible by 42 . |
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Answer» Let `P(n)=n^7-n` Step I For `n=1`. `P(1)=1^7-=0` , which is divisible by 42. Therefore , the result is true for `n=1` . Step II Assume that the result is true for `n=k`. Then , `P(k)=k^7-k` is divisible by 42. `rArr P(k)=42r`, where r is an integer. Step III For `n=k+1`. `P(k+1)=(k+1)^7-(k+1)=(1+k)^7-(k+1)` `=1+.^(7)C_(1)k+.^(7)C_(2)k^2+.^(7)C_(3)k^3+.^(7)C_(4)k^4+.^(7)C_(5)k^5+.^(7)C_(6)k^6+.^(7)C_(7)k^7-(k+1)` `=(k^7-k)+(.^7C_1k+.^7C_2k^2+.^7C_(3)k^3+.^7C_4k^4+.^7C_5+.^(7)C_(6)k^6)` But by assumption `k^7-k` is divisible by 42. Also `.^7C_1k+.^7C_2k^2+.^7C_3k^3+.^7C_4k^4+.^7C_5k^5+.^7C_6k^6` is divisible by 42. `[ because .^7C_r,1 le r le 6 "is divisible by" 7]` Hence , `P(k+1)` is divisible by 42. This shows that , the result is true for `n=k+1`. `therefore` By the principle of mathematical induction , the result is true for all `n in N`. |
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| 5. |
Prove by induction that the integer next greater than `(3+sqrt(5))^n` is divisible by `2^n` for all `n in N`. |
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Answer» Let `alpha=3+sqrt(5) and beta =3-sqrt(5)` `therefore 0 lt beta^(n)lt 1,forall n in N` `rArr alpha +beta=6,alpha beta=4` .....(i) Then , `alpha and beta` are the roots of `x^2-6x+4=0` `rArr alpha^2=6alpha-4` `beta^2=6beta-4` ...(ii) `therefore alpha^2+beta^2=6(alpha+beta)-8=28` ......(iii) `therefore alpha^n+beta^n=(3+sqrt(5))^n+(3-sqrt(5))^n` `=2[3^n+.^(n)C_(2)3^(n-2).5+.^(n)C_(4)3^(n-4).5^(2)+........]` Even integer . As , `0lt beta^(n) lt 1,alpha^(n)+beta^(n)` is the even integer next greater than `alpha^(n)`. Step II For `n=1`, `alpha+beta=6` divisible by `2^1` and `n=2, alpha^2+beta^2=28` divisible by `2^2` which is true for `n=1,2`. Step II Assume it is true for `n=k` . i.e., `alpha^k+beta^k` is divisible by `2^k`. Step III For `n=k+1`. the integer next greater than `alpha^(k+1) is alpha^(k+1)+beta^(k+1)` `= alpha^2.alpha^(k-1)+beta^2.beta^(k-1)` `=(6alpha-4).alpha^(k-1)+(6beta-4).beta^(k-1)` `=6(alpha^k+beta^k)-4(alpha^(k-1)+beta_(k-1))` `=3` (divisible by `2^(k+1))-` (divisible by `2^k+1`) =Divisible by `2^k+1`. This shows that the result is true for `n=k+1`. Hence , the integer next greater than `alpha^(k+1)` is divisible by `2^(k+1)`. |
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| 6. |
For all `n in N, Sigma n`A. `lt((2n+1)^(2))/(8)`B. `gt((2n+1)^(2))/(8)`C. `=((2n+1)^(2))/(8)`D. none of these |
| Answer» Correct Answer - A | |
| 7. |
For all `n in N, 1 + 1/(sqrt(2))+1/(sqrt(3))+1/(sqrt(4))++1/(sqrt(n))`A. equal to `sqrt(n)`B. less than or equal to `sqrt(n)`C. greater tha or equal to `sqrt(n)`D. none of these |
| Answer» Correct Answer - B | |
| 8. |
Statement -1 for all natural numbers n , `2.7^(n)+3.5^(n)-5` is divisible by 24. Statement -2 if f(x) is divisible by x, then `f(x+1)-f(x)` is divisible by `x+1,forall x in N`.A. Statement -1 is true , Statement -2 is true, Statement -2 is correct explanation for Statement -2B. Statement -1 is true , Statement -2 is true , Statement -2 is not correct explanation for Staement -2C. Statement -1 is true , Statement -2 is falseD. Statement -1 is false , Statement - 2 is true. |
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Answer» Let `P(n):2.7^(n)+3.5^(n)-5` Step I For `n=1`, `P(1):2.7^1+3.5^1-5` :24 is divisible by 24. Step II Assume P(k) is divisible by 24, then `P(k):2.7^k+3.5^k-5=24lambda, lambda` is positive integer. Step III For `n=k+1`, `P(k+1)-P(k)=(2.7^(k+1)+3.5^(k+1)-5)-(2.7^k+3.5^k-5)` `=2.7^k(7-1)+3.5^k(5-1)` `12(7^k+5^k)` =divisible by 24 `=24 mu , forall mu in I" "[ because 7^k+5^k "is a always divisible by"24]` `therefore P(k+1)=P(k)+24mu=24lambda+24mu` `=24(lambda+mu)` Hence , `P(k=1)` is divisible by 24. Hence , Statement -1 true and Statement -2 is false . |
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| 9. |
`AA n in N, 49^n+16n-1` is divisible by (A) `64` (B) `49` (C) `132` (D) `32`A. 64B. 8C. 16D. 4 |
| Answer» Correct Answer - A | |
| 10. |
If n is an odd positive integer, then `a^(n)+b^(n)` is divisible byA. a+bB. a-bC. `a^(2)+b^(2)`D. none of these |
| Answer» Correct Answer - A | |
| 11. |
`n(n^2 -1)`, is divisible by 24 if n is an odd positive number.A. 6B. 16C. 36D. 24 |
| Answer» Correct Answer - A | |
| 12. |
For all `n in N`, `4^(n)-3n-1` is divisible byA. 3B. 8C. 9D. 11 |
| Answer» Correct Answer - C | |
| 13. |
For all `n in N, 3^(3n)-26^(n)-1` is divisible byA. 24B. 64C. 17D. 676 |
| Answer» Correct Answer - D | |
| 14. |
If `n in N`, then `3^(2n)+7` is divisible byA. 3B. 8C. 9D. 11 |
| Answer» Correct Answer - B | |
| 15. |
For all `n in N, n^(3)+2n` is divisible byA. 3B. 8C. 9D. 11 |
| Answer» Correct Answer - A | |
| 16. |
`5^(2n)-1` is divisible by `24` for all `n epsilon N `A. 6B. 11C. 24D. 26 |
| Answer» Correct Answer - C | |
| 17. |
For all `n in N, 7^(2n)-48n-1` is divisible byA. 25B. 26C. 1234D. 2304 |
| Answer» Correct Answer - D | |
| 18. |
`1^(3)+2^(2)+3^(3)+. . .+n^(3)=((n(n+1))/(2))^(2)`. |
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Answer» Let `P(n):1^3+2^3+3^3+.....+n^3=[(n(n+1))/(2)]^2`......(i) Step I For `n=1`, LHS of Eq.(i) `=^3=1` and RHS of Eq. (i). `[(1(1+1))/(2)]^2=1^2=1` `therefore LHS=RHS` Therefore ,P(1), is ture. Step II Assume P(k) is true , then `P(k):1^3+2^3+3^3+.....K^3=[(k(k+1))/(2)]^2` Step III For `n=k+1`, `P(k+1):1^3+2^3+3^3+......+^3(k+1)^3` `=[((k+1)+(k+2))/(2)] ^2` LHS `=1^3+2^3+3^3+.....+k^3+(k+1)^3=[(k(k+1))/(2)]^2+(k+1)^3` [by assumption step] `=((k+1)^2)/(4)[k^2+4(k+1)]` `=((k+1)^2(k^2+4k+4))/(4)` `=((k+1)^2(k+2)^2)/(4)` `=[(k+1(k+2))/(2)]^2=RHS` Therefore , `P(k+1)` is true , Hence , by the principle of mathematical induction , P(n)is true for all `n epsi N`. |
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| 19. |
`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+.....+n(n+1)(n+2)=(n(n+1)(n+2)(n+3))/(4)` ........(i) Step I For `n =1, LHS of Eq. (i) `=1.2.3=6`and RHS of Eq. (i) `=(1.(1+1(1+2)(1+3))/(4)=6` `therefore` LHS = RHS Therefore , P(1) is true , then Step II Assume that P(k) is true , then `P(k):1.2.3+2.3.4+.....+k(k+1)(k+2)=(k(k+1)(k+2)(k+3))/(4)` Step III For `n=k+1` `P(k+1):1.2.3+2.3.4+.....+k(k+1)(k+2)+(k+1)(k+2)(k+3)` `=((k+1)(k+2)(k+3)(k+4))/(4)` `therefore LHS = 1.2.3+2.3.4+...+k(k+1)(k+2)+(k+1)(k+2)(k+3)` =(k(k+1)(k+2)(k+3))/(4)+(k+1)(k+2)(k+3)` [by assumption step] `=((k+1)(k+2)(k+3))/(4)(k+4)` `=((k+1)(k+2)(k+3)(k+4))/(4)=RHS` Therefore , `P(k+1)` is true . Hence , by the principle of mathematical induction P(n) is true for all `n epsi N`. |
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| 20. |
Using the principle of mathematical induction, prove that `:``1. 2. 3+2. 3. 4++n(n+1)(n+2)=(n(n+1)(n+2)(n+3))/4^`for all `n in N`. |
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Answer» Let `P(n):(1)/(1.2.3)+(1)/(2.3.4)+.....+(1)/(n(n+1)(n+2))=(n(n+3))/(4(n+1)(n+2))` .....(i) Step I For n=1. LHS is Eq. (i) `=(1)/(1.2.3)=(1)/(6)` and RHS of Eq. (i) `=(1(1+3))/(4(1+1)(1+2))=(1)/(6)` Therefore , P(1) is true . Step II Assume that P(k) is ture , 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))` Step III For `n=k+1`, ``P(k):(1)/(1.2.3)+(1)/(2.3.4)+.....+(1)/(k(k+1)(k+2))+(1)/((k+1)(k+2(k+3)))=((k+1)(k+4))/(4(k+2)(k+3))` `therefore LHS =(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))` [by assumption step ] `=(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+10^2(k+4))/(4(k+1)(k+2)(k+3))` `=((k+1)(k+4))/(4(k+2)(k+3))=RHS` Therefore , `P(k+1)` is true , Hence , by the principle of mathematical P(n) is true for all `n in N`. |
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| 21. |
Prove by mathematical induction that `sum_(r=0)^(n)r^(n)C_(r)=n.2^(n-1), forall n in N`. |
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Answer» Let `P(n):sum_(r=0)^(n)r^(n)C_(r)=n.2^(n-1)` Step I For n=1, LHS of Eq. (i) `=sum_(r=0)^(1)r.^(1)C_(r)=0+1.^(1)C_(1)=1` and RHS of Eq. (i) `=1.2^(1-1)=2^(0)=1` Therefore , P(1) is true . Step II Assume that P(k) is true , then P(k) : `sum_(r=0)^(k)r.^(k)C_(r)=k.2^(k-1)` Step III For `n=k+1` `P(k+1):sum_(r=0)^(k+1)r.^(k+1)C_(r)=(k+1).2^(k)` `therefore LHS =sum_(r=0)^(k+1)r.^(k+1)C_(r)=0+sum_(r=1)^(k+1)r.^(k+1)C_(r)` `=sum_(r=1)^(k+1)r.^(k+1)C_(r)=sum_9r=1)^(k)r.^(k+1)C_(r)+(k+1).^(k+1)C_(k+1)` `=sum_(r=1)^(k)r(.^(k)C_(r)+.^(k)C_(r-1))+(k+1)` `=sum_(r=1)^(k)r.^(k)C_(r)+sum_(r=0)^(k)r.^(k)C_(r-1)+(k+1)` `=sum_(r=0)^(k)r.^(k)C_(r)+sum_(r=0)^(k+1)r.^(k)C_(r-1)` `=sum_(r=0)^(k)r.^(k)C_(r)+sum _(r=0)^(k)(r+1).^(k)C_r)` `=sum_(r=0)^(k)r.^(k)C_(r)+sum_(r=0)^(k)r.^(k)C_(r)+sum_(r=0)^(k).^(k)C_(r)` `=P(k)+P(k)+2^k` [by assumption step] `=k.2^(k-1)+k.26(k-1)+2^(k)=2.k.2^k-1+2^k` `=k.2^k+2^k=(k+1).2^k=RHS` Therefore , `P(k+1)` is true. Hence , by the principle of mathematical induction `P(n)` is true for all `n in N`. |
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| 22. |
If `P(n)=2+4+6+....+2n, n in N`. Then `P(k) =k(k+1)` `rArr P(k+1)=(k+1)(k+2),forall k in N` , So , we can conclude that `P(n)=n(n+1)` forA. all `n in N`B. `n gt 1`C. `n gt 2`D. Nothing can be said |
| Answer» Correct Answer - It is obvious . | |
| 23. |
Using the principle of mathematical induction to show that `tan^(-1)(n+1)x-tan^(-1)x , forall x in N`. |
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Answer» Let `P(n):tan^(-1)((x)/(1+1.2.x^2))+tan^(-1)((x)/(1+2.3.x^2))+.....+tan^(-1)((x)/(1+n(n+1)x^2))` .....(i) `=tan^(-1)(n+1)x-tan^(-1)x` Step I For `n=1`. LHS of Eq. (i) `=tan^(-1)((x)/(1+1.2.x^2))` `=tan^(-1)((2x-x)/(1+2x.x))=tan^(-1)2x-tan^(-1)x` =RHS of Eq.(i) Therefore , P(1) is true. Step II Assume it is true for `n=k`. `P(k):tan^(-1)((x)/(1+1.2x^2))+tan^(-1)((x)/(1+2.3x^2))+......+tan^(-1)((x)/([1+k(k+1)x^2]))` `=tan^(-1)(k+1)x-tan^(-1)x` Step III For `n=k+1`. `P(k+1):tan^(-1)((x)/(1+1.2.x^2))+tan^(-1)((x)/(1+2.3.x^2))+.....+tan^(-1)((x)/(1+k(k+1)x^2))+......+tan^(-1)((x)/(1+(k=1)(k+2)x^2))` `=tan^(-1)(k+2)x-tan^(-1)x` `therefore LHS =tan^(-1)((x)/(1+1.2.x^2))` `+tan^(-1)((x)/(1+2.3.x^2))+....+tan^(-1)((x)/(1+k(k+1)x^2))+tan^(-1)((x)/(1+(k+1)(k+2)x^2))` `tan^(-1)(k+1)x-tan^(-1)x+tan^(-1)((x)/(1+(k+1)(k+2)x^2))` `=tan^(-1)(k+1)x-tan^(-1)x+tan^(-1)(((k+2)x-(k+1)x)/(1+(k+2)x(k+1)x))` `=tan^(-1)(k+1)x-tan^(-1)x+tan^(-1)(k+2)x-tan^(-1)(k+1)x=tan^(-1)(k+2)x-tan^(-1)x=RHS` This shows that the result is true for `n=k+1`. Hence , by the principle of mathematical induction , the result is true for all the `n in N`. |
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| 24. |
If `a_(1)=1,a_(n+1)=(1)/(n+1)a_(n),a ge1`, then prove by induction that `a_(n+1)=(1)/((n+1)!)n in N`. |
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Answer» Let `P(n):a_(n+1)=(1)/((n+1)!),n in N`.....(i) where `a_1=1 and a_(n+1)=(1)/((n+1))a_(n),n ge 1` ......(ii) Step I For n=1, form Eq. (i) , we get `a_(2)=(1)/((1+1)!)=(1)/(2!)` But from Eq. (ii) , we get a_(2)=(1)/((1+1)),a_(1)=(1)/(2)(1)=(1)/(2)` which is true , Also, for n=2 from Eq. (i) we get `a_3=(1)/(3!)=(1)/(6)` But from Eq. (ii) , we get `a_3=(1)/(3),a_2=(1)/(3).(1)/(2)=(1)/(6)` which is also true . Hence ,P(1) and |
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| 25. |
Let `a_(0)=2,a_1=5` and for `n ge 2, a_n=5a_(n-1)-6a_(n-2)`, then prove by induction that `a_(n)=2^(n)+3^(n), forall n ge 0 , n in N`. |
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Answer» Let `P(n) :a_(n)=2^(n)+3^(n),forall n ge 0, n in N`. and `a_(0)=2,a_1=5` and for `nge 0, n in N`. Step I For `n=0`, `a_(0)=2^(0)+3^(0)=1+10=2` which is true as `a_(0)=2` Also , for `n=1,a_(1)=2^(1)+3^(1)=2+3=5` which is also true as `a_1=5` . Hence , `P(0) and P(1)` are true . Step II Assume that `P(k-1) and P(k)` are true. Then , `a_(k-1)=2^(k-1)+3^(k-1)` .......(i) where `a_(k-1)=5a_(k-2)-6a_(k-3) and a_(k)=2^(k)+3^(k)` ........(ii) where `a_(k)=5a_(k-1)-6a_(k-2)` Step III For `n=k+1`. `P(k+1):a_(k+1)=2^(k+1)+3^(k+1), forall k ge 0, k in N`. where `a_(k+1)=5a_(k)-6a_(k-1)` Now, `a_(k+1)=5a_(k)-6a_(k-1)` `=5(2^k+3^k)-6(2^(k-1)+3^(k-1))` [by using Eqs. (i) and (ii)] `=5.2^(k)+5.3^(k)-6.2^(k-1)-6.3^(k-1)` `=2^(k-1)(5.2-6)+3^(k-1)(5.3-6)` `=2^(k-1) 4+3^(k-1).9=2^(k+1)+3^(k+1)` `rArr a_(k+1)=2^(k+1)+3^(k+1)` where `a_(k+1)=5a_(k)-6a_(k-1)` This shows that the result is true for `n=k+1`. Hence by the second principle of mathematical induction , the result is true for `n in N, n ge 0`. |
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| 26. |
if a,b,c,d,e and f are six real numbers such that `a+b+c=d+e+f` `a^2+b^2+c^2=d^2+e^2+f^2` and `a^3+b^3+c^3=d^3+e^3+f^3` , prove by mathematical induction that `a^n+b^n+c^n=d^n+e^n+f^n forall n in N`. |
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Answer» Let `P(n):a^(n)+b^(2)+c^(n)=d^(n)+e^(n)+f^(n),AA n in N " " ..(i)` where `a+b+c+d=e+" "...(ii)` `a^(2)+b^(2)+c=d^(2)+e^(2)+f " "....(iii)` and `a^(2)+b^(3)+c^(3)=d^(3)+e^(2)+^(3)" "...(iv)` Step I from n from Eq. (i) we get `P(1): a+b+c=d+e+f" " ` [ given] Hence the result is true for n 1 Also, for n=2 from Eq(i), we get `P(2): a^(2)+b^(2)+c^(2)=d^(2)+e^(3)+f^(3) " "` [ given] Hennce the result true or n=3 Therefore, P(1) , P(2) and P (3) are true. Step II Assume that `P(k-2),P(k-1)and P(k)` are true, then `P(k-2), a^(k-2)+b^(k-2)= d^(k-2)+e^(k-2)+f^)k-2) " "...(v)` `p(k-1):a^(-1)+b^(k-1)+c^(k-1)=d^(k-1)+e^(k-1)+f^(k-1) " "....(vi)` and `P(k): a^(k)+b^(k)+c^(k)=d^(k)+e^(k)+f^(k) " m "...(vii)` Step III for ` xn=k+1` we shall to prove that `P(k+1):a^(k+1)+b^(k+1)=d^(k+1)+e^(k+1)+f^(k+1)` LHS `=a^(k+1)+b^(k+1)+c^(k+1)` `=(a^(k)+b^(k)(a+b+c)-(a^(k-1)+b^(k-1)+c^(k-1))` `(ab+bc+ca)+abc(a^(k-2)+b^(k-2)+b^(k-2)+c^(k-2))` `=(d^(k)+e^(f)+f^(k))(d+e+f)-(d^(k-1)+e^(k-1)+c^(k-2))` `(de+ef+fd)+def(d^(k-2)+e^(k-2)+f^(k-2))` [ using Eqs. (ii), (iii), (iv), (v), (vi), (vii)] `:. (a+b+c)^(2)=(d+e+f)^(2)` `rArr a^(2)+b^(2)+c^(2)+2(ab+bc+ca)` `=d^(2)+e^(2)+f^(2)+2(de+ef+fd)` `rArr ab+bc+ca=de+ef+fd` `[ :. a^(2)+b^(2)+c^(2)=d^(2)+e^(2)+f^(2)]` and `a^(3)+b^(3)+c^(3) -3abc` `(=d+e+f)(d^(2)+e^(2)+f^(2)-de-ef-fd)` `=d^(3)+e^(3)+f^(3)-edf` `rArr abc=def [ :. a^(3)+b^(3)+c^(3)=d^(3)+e^(3)+f^(3))` `=d^(k+1)+e^(k+1)+f^(k-1)=RHS` This shows that result is true for n=k+1. Hence by second perincipal of mathmatical inducition, the result is true for all `n in N`. |
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| 27. |
`1^2/(1.3)+2^2/(3.5)+3^2/(5.7)+.....+n^2/((2n-1)(2n+1))=((n)(n+1))/((2(2n+1))` |
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Answer» Let `P(n):(1^2)/(1:3)+(2^2)/(3.5)+.....+(n^2)/((2n-1)(2n+1))=(n(n+1))/(2(2n+1))` .....(i) Step I For n=1. LHS of Eq. (i) `(1^2)/(1.3)=(1)/(3)` RHS of Eq. (i) `=(1(1+1))/(2(2xx1+1))=(2)/(2(3))=(1)/(3)` LHS=RHS Therefore , P(1) is true. Step II Let us assume that the result is true for `n=k`, then `P(k)=(1^2)/(1.3)+(2^2)/(3.5)+....+(k^2)/((2k-1)(2k+1))=(k(k+1))/(2(2k+1))` Step III For `n=k+1`, we have to prove that `P(k+1):(1^2)/(1:3)+(2^2)/(3.5)+......+(k^2)/((2k-1)(2k+1))+((k+1)^2)/((2k+1)(2k+3))=((k+1)(k+2))/(2(2k+3))` LHS `=(1^2)/(1.3)+(2^2)/(3.5)+......+(k^2)/((2k-1)(2k+1))+((2k+1)^2)/((2k+1)(2k+1))` `=(k(k+1))/(2(2k+1))+((k+1)^2)/((2k+1)(2k+3))` [ by assumption step] `=((k+1))/((2k+1)){(k)/(2)+(k+1)/((2+3))}=((k+1))/((2k+1)){(2k^2+5k+2)/(2(2k+3))}` `=((k+1))/((2k+1)).((k+2)(2k+1))/(2(2k+3))=((k+1)(k+2))/(2(2k+3))=RHS` This shows that ,the result is true for `n=k+1`. Therefore , by the principle of mathematical induction the result is true for all ` n in N`. |
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| 28. |
Prove that `tan^(- 1)(1/3)+tan^(- 1)(1/7)+tan^(- 1)(1/13)+..........+tan^-1 (1/(n^2+n+1))+......oo =pi/4` |
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Answer» Let `P(n):tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+…..+tan^(-1)((1)/(n^+n+1))=tan^(-1)((n)/(n+2))` ……..(i) Step I For `n=1` LHS of Eq. (i) `=tan^(-1)((1)/(3)) =tan^(-1)((1)/(1+2))=RHS of Eq. (i) Therefore , P(1) is true . Step II Assume that P(k) is true. Then , `P(k):tan^(-1)((1)/(3))+tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+........+tan^(-1)((1)/(k^2+k+1))=tan^(-1)((k)/(k+2))` Step III For `n=k+1` `P(k+1):tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+....+tan^(-1)((1)/(k^2+k+1))+tan^(-1)((1)/((k+1)^2+(k+1)+1))=tan^(-1)((k+1)/(k+3))`......(ii) LHS of Eq. (ii) `=tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+.....+tan^(-1)((1)/(k^2+k+1))+tan^(-1)((1)/((k+1)^2+(k+1)+1))` `=tan^(-1)((k+1)/(k+2))+tan^(-1)((1)/((k+1)^2+(k+1)+1))` [by aaumption atep] `=tan^(-1)((k)/(1+(k+1)))+tan^(-1)((1)/(k^2+3k+3))` `=tan^(-1)((k)/(1+(k+1)))+tan^(-1)((1)/(1+(k+1)(k+2)))` `=tan^(-1)(((k+1)-1)/(1+(k+1).1))+tan^(-1)(((k+2)-(k+1))/(1+(k+2)(k+1)))` `=tan^(-1)(k+1)-tan^(-1)1+tan^(-1)(k+2)-tan^(-1)(k+1)=tan^(-1)(k+2)-tan^(-1)1` `=tan^(-1)((k+2-1)/(1+(k+2).1))=tan^(-1)((k+1)/(k+3))=` RHS of Eq. (ii) This shows that the result is true for `n=k+1`. Hence. by the principle of mathematical induction . the result is true for all `n in N`. |
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| 29. |
3 + 13 + 29 + 51 + 79+… to n terms =A. `2n^(2)+7n^(3)`B. `n^(2)+5n^(3)`C. `n^(3)+2n^(2)`D. none of these |
| Answer» Correct Answer - C | |
| 30. |
`n^(th)` term of the series `4+14+30+52+ .......=`A. 5n - 1B. `2n^(2) + 2n`C. `3n^(2)+n`D. `2n^(2)+2` |
| Answer» Correct Answer - C | |
| 31. |
For all `n in N, (n^(5))/(5)+(n^(3))/(3)+(7)/(15n)` isA. an integerB. a natural numberC. a positive fractionD. none of these |
| Answer» Correct Answer - B | |
| 32. |
If 3+5+9+17+33+… to n terms `=2^(n+1)+n-2`, then nth term of LHS, isA. `3^(n)-1`B. `2n+1`C. `2^(n)+1`D. `3n-1` |
| Answer» Correct Answer - C | |
| 33. |
The sum of n terms of the series `1+(1+a)+(1+a+a^(2))+(1+a+a^(2)+a^(3))+...,` isA. `(n)/(1-a)-(a(1-a^(n)))/((1-a)^(2))`B. `(n)/(1-a)+(a(1-a^(n)))/((1-a)^(2))`C. `(n)/(1-a)+(a(1+a^(n)))/((1-a)^(2))`D. `-(n)/(1-a)+(a(1-a^(n)))/((1-a)^(2))` |
| Answer» Correct Answer - A | |
| 34. |
Find the sum of the first n terms of the series : `3 + 7 + 13 +21 + 31 +dot dot dot`A. 4n-1B. `n^(2)+2n`C. `n^(2)+n+1`D. `n^(2)+2` |
| Answer» Correct Answer - C | |
| 35. |
Prove `1.4.7+2.5.8+3.6.9+.......` upto n terms `=(n)/(4)(n+1)(n+6)(n+7)` |
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Answer» Let `P(n):1.4.7+2.5.8+3.6.9+......+` upto n terms `(n)/(4)(n+1)(n+6)(n+7)` i.e., `P(n):1.4.7+2.5.8+3.6.9+......+n(n+3)(n+6)=(n)/(4)(n+1)(n+6)(n+7)` Step I For `n=1`, LHS of Eq. (i) `=1.4.7=28` RHS of Eq. (i) `=(1)/(4)(1+1)(1+6)(1+7)=(2.7.8)/(4)=28` LHS = RHS Therefore, P(1) is true . Step II Let us assume that the result is true for `n=k`. Then , `P(k):1.4.7+2.5.8+3.6.9+......+k(k+3)(k+6)=(k)/(4)(k+1)(k+6)(k+7)` Step III For `n=k+1`, we have to prove that `P(k+1):1.4.7+2.5.8+3.6.9+.....+k(k+3)(K+6)+(k+1)(k+4)(k+7)` `=((k+1))/(4)(k+2)(k+7)(k+8)` LHS =1.4gt7+2.5.8+3.6.9+......+k(k+3)(k+6)+(k+1)(k+4)(k+7)` =(k)/(4)(k+1)(k+6)(k+7)+(k+1)(k+4)(k+7)`[by assumption step ] `=(k+1)(k+7){(k)/(4)(k+6)+(k+4)}` `=(k+1)(k+7){(k^2+6k+4k+16)/(4)}` `=(k+1)(k+7){(k^2+10k+16)/(4)}` `=(k+1)(k+7){((k+2)(k+8))/(4)}` `=((k+1))/(4)(k+2)(k+7)(k+8)=RHS` This shows that the result is true for `n=k+1`. Hence, by the principle of mathematical induction , the result is true for all `n in N`. |
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| 36. |
If n is an even positive integer, then `a^(n)+b^(n)` is divisible byA. a+bB. a-bC. `a^(2)-b^(2)`D. none of these |
| Answer» Correct Answer - D | |
| 37. |
Sum of n terms of the following series`1^3+3^3+5^3+7^3+`A. `n^(2)(n^(2)-1)`B. `n^(2)(2n^(2)-1)`C. `n^(2)(2n^(2)+1)`D. `n^(2)(n^(2)+1)` |
| Answer» Correct Answer - B | |
| 38. |
If a,b are distinct rational numbers, then for all `n in N` the number `a^(n)-b^(n)` is divisible byA. a-bB. a+bC. 2a-bD. a-2b |
| Answer» Correct Answer - A | |
| 39. |
If `10^n+3xx4^(n+2)+lambda`is divisible by 9 or all `n N`, then the least positive integral value of `lambda`is`5`b. `3`c. `7`d. `1`A. 5B. 3C. 7D. 1 |
| Answer» Correct Answer - A | |
| 40. |
If `x^n-1` is divisible by `x-k` then the least positive integral value of k isA. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - A | |
| 41. |
`1.3+3.5+5.7+......+(2n-1)(2n+1)=(n(4n^2+6n-1))/3` |
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Answer» Let `P9n):1.3+3.5+5.7+....+(2n-1)(2n+1)` `(n(4n^2+6n-1))/(3)` Step I For `n=1`. LHS of Eq. (i) `=1.3=3` RHS of Eq. (i) `=(1(4xx1^2+6xx1-1))/(3)=(4+6-1)/(3)=3` `therefore` LHS = RHS Therefore , P(1) is true. Step II Assume that the result is true for `n=k`. Then, `P(k):1.3+3.5+5.7+.....+(2k-1)(2k+1)=(k(4k^2+6k-1))/(3)` Step III For `n=k+1`, we have to prove that `P(k+1):1.3+3.5+5.7+....+(2k-1)(2k+1)+(2k+1)(2k+3)` `=((k+1)[4(k+1)^2+6(k+1)-1])/(3)` `=((k+1)(4k^2+14k+9))/(3)` LHS `=1.3+3.5+5.7+...+(2k-1)(2k+1)+(2k+1)(2k+3)` `=(k(4k^2+6k-1))/(3)+(2k+1)(2k+3)` [by assumpation step ] `=(4k^3+6k^2-k)/(3)+(4k^2+8k+3)` `=(4k^3+18k^2+23k+9)/(3)` `=((k+1)(4k^2+14k+9))/(3)=RHS` This shows that the result is true for `n=k+1`. Therefore , by the principle of mathematical induction , the result is true for all `n in N`. |
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| 42. |
When the square of any odd number, greater than1, is divided by 8, it always leaves remainder1(b) 6 (c) 8 (d) Cannot be determined |
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Answer» Let `P(n):(2r+1)^(2n), forall n in N and r in I`. Step I For `n=1`. `P(1):(2r+1)^2=4r^2+4r+1=4r(r+1)+1=8p+1,p in I " "[because r(r+1)"is an even integer"]` Therefore , `P(1)` is true , Step II Assume P(n) is true for n=k , then `P(k):(2r+1)^2k` is divisible by 8 levaes remainder 1. `rArr P(k)=8m+1,n in I`, where m is a positive integer . Step III For `n=k+1`. brgt `therefore P(k)=(2r+1)2(k+1)` `=(2r+1)^(2k)(2r+1)^2` `=(8m+1)(8p+1)` `64mp+8(m+p)+1` `=8(8mp+m+p)+1` which is true for `n=k+1` as `8mp+m+p` is an integer. Hence , by the principle of mathematical induction, when P(n) is divided by 8 leaves the ramainder 1 for all `n in N`. |
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| 43. |
`1^2+2^2+3^2++n^2=(n(n+1)(2n+1))/6` |
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Answer» Let `P(n):1^2_2^2+3^2+....+n^2=((n+1)(2n+1))/(6)` Step I For n=1 , LHS of Eq. (i) `=1^2=1` RHS of Eq. (i) `=((1+1)(2xx1+1))/(6)` `=(1.2.3)/(6)=1` LHS = RHS Therefore , P(1) is true . Step II Let us assume that the result is true for `n=k`. Then , `P(k):1^2+2^2+3^2+......+k^2=(k(k+1)(2k+1))/(6)` Step III For `n=k+1`, we have to prove that `P(k+1):1^2+2^2+3^2+......+k^2+(k+1)^2` `=((k+1)(k+2)(k+3))/(6)` LHS =`1^2+2^2+3^2+....+k^2+(k+1)^2` `=(k(k+1)(2k+1))/(6)+(k+1)^2` `=(k(k+1)(2k+1))/(6)+(k+1)^2` `=(k+1){(k(2k+1))/(6)+(k+1)}` `=(k+1){(2k^2+7k+6)/(6)}` `=(k+1){((k+2)(2k+3))/(6)}=((k+1)(k+2)(2k+3))/(6)= RHS ` This shows that the result is true for `n=k+1`. Therefore , by the principle of methematical induction ,the result is true for all `n in N`. |
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| 44. |
Prove the following by using iduction for all `n in N`. `1+2+3+.....+n=(n(n+1))/(2)` |
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Answer» Let `P(n):1+2+3+....+n(n(n+1))/(2)=1` Step I For `n=1`, LHS of Eq. (i) =1 RHS of Eq. 9i) `=(1(1+1))/(2)=1` LHS = RHS Therefore , P(1) is true. Step II Let us assume that the result is true for `n=k` , Then , P(k) `: 1 + 2 + 3 + ... + k = (k(k+ 1))/(2)` Step III For n = k + 1 , we have to prove that P(k+ 1) = ` 1 + 2 + 3 + ... + k + (k + 1) = ((k + 1) (k + 2))/(2)` L.H.S = 1 + 2 + 3 + ... + k+ ( k + 1) ` = (k(k + 1))/(2) + k + 1 ` [ By assumption step] `= (k + 1) ((k)/(2) + 1) = (k + 1) ((k+ 2)/(2))` ` = (( k + 1 ) (k + 2))/(2)` = RHS This show that the result is true for n = k + 1 . Therefore , by the principle of mathematical induction , the result is true for all `n in N` |
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