Q44 / 160 · TS EAMCET 2004 · Mathematics

$\sum_{k=1}^5 \frac{1^3+2^3+\ldots+k^3}{1+3+5+\ldots+(2 k-1)}$ is equal to

  • Correct answer
    A
    $22.5$
  • B
    $24.5$
  • C
    $28.5$
  • D
    $32.5$

Solution

$\begin{aligned} & \sum_{k=1}^5 \frac{1^3+2^3+\ldots+k^3}{1+3+5+\ldots .+(2 k-1)} \\ & =\sum_{k=1}^5 \frac{\left(\frac{k(k+1)}{2}\right)^2}{k^2} \\ & \qquad\left[\begin{array}{l}\because n^3=\left(\frac{n(n+1)}{2}\right)^2 \\ \text { and } 1+3+5 \ldots+(2 K-1)=K^2\end{array}\right] \\ & =\sum_{k=1}^5 \frac{(k+1)^2}{4}=\frac{2^2+3^2+4^2+5^2+6^2}{4} \\ & =\frac{4+9+16+25+36}{4}=\frac{90}{4}=22.5\end{aligned}$

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

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