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TS EAMCET · Maths · Sequences and Series

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

  1. A \(22.5\)
  2. B \(24.5\)
  3. C \(28.5\)
  4. D \(32.5\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(22.5\)

Step-by-step Solution

Detailed explanation

\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…