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

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(A) Let the given sum be $S$. The $n^{th}$ term of the series is $T_n = \frac{\sum_{k=1}^n k^3}{\sum_{k=1}^n k}$.Using the formulas $\sum_{k=1}^n k^3

Vedclass · Accepted Answer

$620$

Vedclass · Answer

(A) Let the given sum be $S$. The $n^{th}$ term of the series is $T_n = \frac{\sum_{k=1}^n k^3}{\sum_{k=1}^n k}$.Using the formulas $\sum_{k=1}^n k^3 = \left[\frac{n(n+1)}{2}ight]^2$ and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$,we get $T_n = \frac{[n(n+1)/2]^2}{n(n+1)/2} = \frac{n(n+1)}{2}$.The sum of the first $15$ terms is $\sum_{n=1}^{15} T_n = \sum_{n=1}^{15} \frac{n(n+1)}{2} = \frac{1}{2} \left[ \sum_{n=1}^{15} n^2 + \sum_{n=1}^{15} n ight]$.Using $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum n = \frac{n(n+1)}{2}$,for $n=15$:Sum $= \frac{1}{2} \left[ \frac{15 	imes 16 	imes 31}{6} + \frac{15 	imes 16}{2} ight] = \frac{1}{2} [1240 + 120] = \frac{1360}{2} = 680$.The expression is $\sum_{n=1}^{15} T_n - \frac{1}{2} \sum_{n=1}^{15} n = 680 - \frac{1}{2} 	imes 120 = 680 - 60 = 620$.

$1 + \frac{1^3 + 2^3}{1 + 2} + \frac{1^3 + 2^3 + 3^3}{1 + 2 + 3} + \dots + \frac{1^3 + 2^3 + 3^3 + \dots + 15^3}{1 + 2 + 3 + \dots + 15} - \frac{1}{2}(1 + 2 + 3 + \dots + 15)$ is equal to

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