If 4^3+8^3+12^3+ldots up to n terms = k n^2(n | vedclass

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(C) The given series is $S_n = 4^3 + 8^3 + 12^3 + \ldots + (4n)^3$. This can be written as $S_n = \sum_{r=1}^{n} (4r)^3$. $S_n = \sum_{r=1}^{n} 64r^3

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) The given series is $S_n = 4^3 + 8^3 + 12^3 + \ldots + (4n)^3$. This can be written as $S_n = \sum_{r=1}^{n} (4r)^3$. $S_n = \sum_{r=1}^{n} 64r^3 = 64 \sum_{r=1}^{n} r^3$. Using the formula for the sum of cubes of the first $n$ natural numbers,$\sum_{r=1}^{n} r^3 = \left[ \frac{n(n+1)}{2} ight]^2 = \frac{n^2(n+1)^2}{4}$. Substituting this into the expression for $S_n$: $S_n = 64 	imes \frac{n^2(n+1)^2}{4} = 16n^2(n+1)^2$. Comparing this with the given form $k n^2(n+1)^2$,we get $k = 16$.

If $4^3+8^3+12^3+\ldots$ up to $n$ terms $= k n^2(n+1)^2$ (for all $n \in N$),then $k=$

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