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

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(C) The $n^{th}$ term $T_n$ is given by the sum of the first $n$ cubes in the numerator and the sum of the first $n$ odd numbers in the denominator.Nu

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The $n^{th}$ term $T_n$ is given by the sum of the first $n$ cubes in the numerator and the sum of the first $n$ odd numbers in the denominator.Numerator: $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2 = \frac{n^2(n+1)^2}{4}$.Denominator: The sum of the first $n$ odd numbers is $1 + 3 + 5 + \dots + (2n-1)$. This is an arithmetic progression with $a=1$,$d=2$,and $n$ terms. The sum is $\frac{n}{2}[2(1) + (n-1)2] = \frac{n}{2}[2 + 2n - 2] = n^2$.Therefore,$T_n = \frac{\frac{n^2(n+1)^2}{4}}{n^2} = \frac{(n+1)^2}{4} = \frac{n^2 + 2n + 1}{4}$.

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

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