If the sum of $n$ terms of the series $2^{3} + 4^{3} + 6^{3} + ...$ is 3528, then $n$ equals to

Question

Dr. Abubacker · Accepted Answer

(d)
Given series is $2^{3} + 4^{3} + 6^{3} + ...$
$	herefore T_{n} = (2n)^{3} = 8n^{3}$
$	herefore\Sigma T_{n} = 8\Sigma n^{3} = \frac{8n^{2}(n + 1)^{2}}{4}$
$\Rightarrow 3528 = 2n^{2}(n + 1)^{2} \Rightarrow n = 6$

Q. If the sum of $n$ terms of the series $2^{3} + 4^{3} + 6^{3} + ...$ is 3528, then $n$ equals to

Explanation

Reviewed by Dr. Abubacker

Want more practice?