The {n^{th}} term of the series frac{1^3}{1} | vedclass

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(C) The ${n^{th}}$ term is given by $T_n = \frac{\sum_{k=1}^{n} k^3}{\sum_{k=1}^{n} (2k - 1)}$.The numerator is the sum of cubes of the first $n$ natu

Vedclass · Accepted Answer

$\frac{n^2 + 2n + 1}{4}$

Vedclass · Answer

(C) The ${n^{th}}$ term is given by $T_n = \frac{\sum_{k=1}^{n} k^3}{\sum_{k=1}^{n} (2k - 1)}$.The numerator is the sum of cubes of the first $n$ natural numbers: $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2 = \frac{n^2(n+1)^2}{4}$.The denominator is the sum of the first $n$ odd numbers,which is $n^2$: $\sum_{k=1}^{n} (2k - 1) = n^2$.Therefore,$T_n = \frac{n^2(n+1)^2 / 4}{n^2} = \frac{(n+1)^2}{4} = \frac{n^2 + 2n + 1}{4}$.

The ${n^{th}}$ term of the series $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots$ is:

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