The sum of the first 9 terms of the series fr | vedclass

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(C) The $n^{th}$ term of the series is given by $T_n = \frac{\sum_{k=1}^n k^3}{\sum_{k=1}^n (2k-1)}$.We know that $\sum_{k=1}^n k^3 = \left[\frac{n(n+

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$96$

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(C) The $n^{th}$ term of the series is given by $T_n = \frac{\sum_{k=1}^n k^3}{\sum_{k=1}^n (2k-1)}$.We know that $\sum_{k=1}^n k^3 = \left[\frac{n(n+1)}{2}\right]^2$ and the sum of the first $n$ odd numbers is $\sum_{k=1}^n (2k-1) = n^2$.Thus,$T_n = \frac{\left[\frac{n(n+1)}{2}\right]^2}{n^2} = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4}$.We need to find the sum of the first $9$ terms,$S_9 = \sum_{n=1}^9 T_n = \sum_{n=1}^9 \frac{(n+1)^2}{4}$.$S_9 = \frac{1}{4} \sum_{n=1}^9 (n+1)^2 = \frac{1}{4} (2^2 + 3^2 + \dots + 10^2)$.Adding and subtracting $1^2$,we get $S_9 = \frac{1}{4} [(\sum_{k=1}^{10} k^2) - 1^2]$.Using the formula $\sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6}$ for $m=10$:$S_9 = \frac{1}{4} [\frac{10(11)(21)}{6} - 1] = \frac{1}{4} [385 - 1] = \frac{384}{4} = 96$.

The sum of the first $9$ terms 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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