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
- A \(22.5\)
- B \(24.5\)
- C \(28.5\)
- D \(32.5\)
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…
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