The sum frac{1^3}{1} + frac{1^3 + 2^3}{1 + 3} | vedclass

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(B) The $n$-th term is $T_n = \frac{\sum_{i=1}^n i^3}{\sum_{i=1}^n (2i-1)}$.We know that $\sum_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$ and $\sum_{i=1}^n (

Vedclass · Accepted Answer

$71$

Vedclass · Answer

(B) The $n$-th term is $T_n = \frac{\sum_{i=1}^n i^3}{\sum_{i=1}^n (2i-1)}$.We know that $\sum_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$ and $\sum_{i=1}^n (2i-1) = n^2$.Thus, $T_n = \frac{(\frac{n(n+1)}{2})^2}{n^2} = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4} = \frac{n^2+2n+1}{4}$.The sum of $8$ terms is $S_8 = \sum_{n=1}^8 \frac{n^2+2n+1}{4} = \frac{1}{4} [ \sum_{n=1}^8 n^2 + 2\sum_{n=1}^8 n + \sum_{n=1}^8 1 ]$.Using the formulas $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum n = \frac{n(n+1)}{2}$:$S_8 = \frac{1}{4} [ \frac{8(9)(17)}{6} + 2(\frac{8(9)}{2}) + 8 ] = \frac{1}{4} [ 204 + 72 + 8 ] = \frac{284}{4} = 71$.

The sum $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots$ up to $8$ terms, is:

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