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Question: Answered & Verified by Expert
If $\quad 2^3+4^3+6^3+\ldots+2(n)^3=h n^2(n+1)^2$, then $h$ is equal to
MathematicsSequences and SeriesTS EAMCETTS EAMCET 2001
Options:
  • A $\frac{1}{2}$
  • B $1$
  • C $\frac{3}{2}$
  • D $2$
Solution:
2029 Upvotes Verified Answer
The correct answer is: $2$
$\begin{aligned} \text { Let } S_n=2^3+4^3 & +6^3+\ldots+(2 n)^3 \\ & =\Sigma(2 n)^3=8 \Sigma n^3=8 \quad\left[\frac{n(n+1)}{2}\right]^2 \\ \Rightarrow 2 n^2(n+1)^2 & =h n^2(n+1)^2 \\ \Rightarrow \quad h & =2\end{aligned}$

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