Sum to n terms of the series (14/1.3)+(24/3.5)+(34/5.7)+..... is equal to

Question

Sum to n terms of the series (14/1.3)+(24/3.5)+(34/5.7)+..... is equal to (A) (n (n + 1) (2 n2 + 1)/8 (2 n + 1)) (B) (n (n + 1) (n2 + n + 1)/6 (2 n + 1)) (C) ((n + 1) ((2 n + 1)2 + 1)/8 (2 n + 1)) (D) (n (n + 1) ((2 n + 1)2 + 1)/16 (2 n + 1))

Tardigrade Admin · Accepted Answer

The kt h term of the given series is uk=(k4/(2 k - 1) (2 k + 1)) =(k2/4)+(1/16)+(1/32)((1/2 k - 1) - (1/2 k + 1)) Therefore the sum of n terms is Sn=u1+u2+....+un =(1/4) displaystyle ∑ k = 1n k2 + (n/16)+(1/32)(1 - (1/2 n + 1)) =(n (n + 1) (2 n + 1)/4.6)+(n/16)+(n/16 (2 n + 1)) =(n (n + 1) (n2 + n + 1)/6 (2 n + 1))

Q. Sum to $n$ terms of the series $\frac{1 ^{4}}{1.3} + \frac{2 ^{4}}{3.5} + \frac{3 ^{4}}{5.7} + .....$ is equal to

$\frac{n ( n + 1 ) ( 2 n ^{2} + 1 )}{8 ( 2 n + 1 )}$

$\frac{n ( n + 1 ) ( n ^{2} + n + 1 )}{6 ( 2 n + 1 )}$

$\frac{( n + 1 ) ( ( 2 n + 1 ) ^{2} + 1 )}{8 ( 2 n + 1 )}$

$\frac{n ( n + 1 ) ( ( 2 n + 1 ) ^{2} + 1 )}{16 ( 2 n + 1 )}$

Q. Sum to n terms of the series 1.314​+3.524​+5.734​+..... is equal to

8(2n+1)n(n+1)(2n2+1)​

6(2n+1)n(n+1)(n2+n+1)​

8(2n+1)(n+1)((2n+1)2+1)​

16(2n+1)n(n+1)((2n+1)2+1)​

The kth term of the given series is uk​=(2k−1)(2k+1)k4​ =4k2​+161​+321​(2k−11​−2k+11​) Therefore the sum of n terms is Sn​=u1​+u2​+....+un​ =41​k=1∑n​k2+16n​+321​(1−2n+11​) =4.6n(n+1)(2n+1)​+16n​+16(2n+1)n​ =6(2n+1)n(n+1)(n2+n+1)​

Q. Sum to $n$ terms of the series $\frac{1 ^{4}}{1.3} + \frac{2 ^{4}}{3.5} + \frac{3 ^{4}}{5.7} + .....$ is equal to

$\frac{n ( n + 1 ) ( 2 n ^{2} + 1 )}{8 ( 2 n + 1 )}$

$\frac{n ( n + 1 ) ( n ^{2} + n + 1 )}{6 ( 2 n + 1 )}$

$\frac{( n + 1 ) ( ( 2 n + 1 ) ^{2} + 1 )}{8 ( 2 n + 1 )}$

$\frac{n ( n + 1 ) ( ( 2 n + 1 ) ^{2} + 1 )}{16 ( 2 n + 1 )}$