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JEE Mains · Maths · STD 11 - 8. sequence and series

If the sum  \(\frac{3}{1^2} + \frac{5}{{{1^2} + {2^2}}} + \frac{7}{{{1^2} + {2^2} + {3^2}}} + ...... + \) up to \(20\) terms is equal to \(\frac{k}{{21}}\), then \(k\) is equal to

  1. A \(120\)
  2. B \(180\)
  3. C \(240\)
  4. D \(60\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(120\)

Step-by-step Solution

Detailed explanation

\({n^{th}}\) term of given series is \(\frac{{2n + 1}}{{\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}} = \frac{6}{{n\left( {n + 1} \right)}}\) Let \({n^{th}}\) term, \({a_n} = 6\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right]\) Sum of \(20\) terms,…
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