If 2^3+4^3+6^3+ldots+(2n)^3 = h n^2(n+1)^2,th | vedclass

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(D) The given series is $S_n = 2^3 + 4^3 + 6^3 + \ldots + (2n)^3$. We can write this as $S_n = \sum_{k=1}^{n} (2k)^3$. $S_n = \sum_{k=1}^{n} 8k^3 = 8

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) The given series is $S_n = 2^3 + 4^3 + 6^3 + \ldots + (2n)^3$. We can write this as $S_n = \sum_{k=1}^{n} (2k)^3$. $S_n = \sum_{k=1}^{n} 8k^3 = 8 \sum_{k=1}^{n} k^3$. Using the formula for the sum of cubes of the first $n$ natural numbers,$\sum_{k=1}^{n} k^3 = [\frac{n(n+1)}{2}]^2$. Thus,$S_n = 8 	imes [\frac{n(n+1)}{2}]^2 = 8 	imes \frac{n^2(n+1)^2}{4} = 2n^2(n+1)^2$. Comparing this with the given expression $h n^2(n+1)^2$,we get $h = 2$.

If $2^3+4^3+6^3+\ldots+(2n)^3 = h n^2(n+1)^2$,then $h$ is equal to

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