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JEE Mains · Maths · STD 11 - 8. sequence and series
The sum \(\frac{3}{{{1^2}}} + \frac{5}{{{1^2} + {2^2}}} + \frac{7}{{{1^2} + {2^2} + {3^2}}} + ....\) upto \(11-\) terms is
- A \(\frac {7}{2}\)
- B \(\frac {11}{4}\)
- C \(\frac {11}{2}\)
- D \(\frac {60}{11}\)
Answer & Solution
Correct Answer
(C) \(\frac {11}{2}\)
Step-by-step Solution
Detailed explanation
Given sum is \(\frac{3}{{12}} + \frac{5}{{{1^2} + {2^2}}} + \frac{7}{{{1^2} + {2^2} + {3^2}}} + .....\) \({n^{th}}\) term \( = {T_n}\) \( = \frac{{2n + 1}}{{\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}} = \frac{6}{{n\left( {n + 1} \right)}}\) or…
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