The sum of the series: (2)^2 + 2(4)^2 + 3(6)^ | vedclass

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(C) The given series is $S = 1(2)^2 + 2(4)^2 + 3(6)^2 + \dots + 10(20)^2$.The $n^{th}$ term is $T_n = n(2n)^2 = n(4n^2) = 4n^3$.The sum of $10$ terms

Vedclass · Accepted Answer

$12100$

Vedclass · Answer

(C) The given series is $S = 1(2)^2 + 2(4)^2 + 3(6)^2 + \dots + 10(20)^2$.The $n^{th}$ term is $T_n = n(2n)^2 = n(4n^2) = 4n^3$.The sum of $10$ terms is $S_{10} = \sum_{n=1}^{10} 4n^3$.$S_{10} = 4 \sum_{n=1}^{10} n^3$.Using the formula $\sum_{n=1}^{k} n^3 = \left( \frac{k(k+1)}{2} ight)^2$,we get:$S_{10} = 4 \left( \frac{10 	imes 11}{2} ight)^2$.$S_{10} = 4 	imes (55)^2 = 4 	imes 3025 = 12100$.

The sum of the series: $(2)^2 + 2(4)^2 + 3(6)^2 + \dots$ up to $10$ terms is

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