Question:
The sum \(\dfrac{3}{1^2} + \dfrac{5}{1^2+2^2} + \dfrac{7}{1^2+2^2+3^2} + \cdots\) upto 11-terms is :
The sum \(\dfrac{3}{1^2} + \dfrac{5}{1^2+2^2} + \dfrac{7}{1^2+2^2+3^2} + \cdots\) upto 11-terms is :
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Write the general term using \(1^2 + 2^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6}\) and telescope.
Updated On: Oct 31, 2026
- \(\dfrac{7}{2}\)
- \(\dfrac{11}{4}\)
- \(\dfrac{11}{2}\)
- \(\dfrac{60}{11}\)
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The Correct Option is C
Solution and Explanation
The \(k\)-th term is \(T_k = \dfrac{2k+1}{1^2 + 2^2 + \cdots + k^2} = \dfrac{2k+1}{\frac{k(k+1)(2k+1)}{6}} = \dfrac{6}{k(k+1)}\).
\(T_k = 6\left(\dfrac{1}{k} - \dfrac{1}{k+1}\right)\).
\(S_{11} = 6\left(1 - \dfrac{1}{12}\right) = 6\cdot\dfrac{11}{12} = \dfrac{11}{2}\).
Answer: (3) \(\dfrac{11}{2}\)
\(T_k = 6\left(\dfrac{1}{k} - \dfrac{1}{k+1}\right)\).
\(S_{11} = 6\left(1 - \dfrac{1}{12}\right) = 6\cdot\dfrac{11}{12} = \dfrac{11}{2}\).
Answer: (3) \(\dfrac{11}{2}\)
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