Q117 / 200 · TS EAMCET 2001 · Mathematics

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

  • A
    $\frac{1}{2}$
  • B
    $1$
  • C
    $\frac{3}{2}$
  • Correct answer
    D
    $2$

Solution

$\begin{aligned} \text { Let } S_n=2^3+4^3 & +6^3+\ldots+(2 n)^3 \\ & =\Sigma(2 n)^3=8 \Sigma n^3=8 \quad\left[\frac{n(n+1)}{2}\right]^2 \\ \Rightarrow 2 n^2(n+1)^2 & =h n^2(n+1)^2 \\ \Rightarrow \quad h & =2\end{aligned}$

Hence, the correct answer is option: 'D'.

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