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

योगफल \(1+\frac{1^{3}+2^{3}}{1+2}+\frac{1^{3}+2^{3}+3^{3}}{1+2+3}+\ldots\) \(+\frac{1^{3}+2^{3}+3^{3}+\ldots .+15^{3}}{1+2+3+\ldots .+15}-\frac{1}{2}(1+2+3+\ldots .+15)\) बराबर है 

  1. A \(620\)
  2. B \(1860\)
  3. C \(1240\)
  4. D \(660\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(620\)

Step-by-step Solution

Detailed explanation

Sum \( = \sum\limits_{n = 1}^{15} {\frac{{{1^3} + {2^3} + ...{n^3}}}{{1 + 2 + ... + n}}} - \frac{1}{2}.\frac{{15.16}}{2}\) \( = \sum\limits_{n = 1}^{15} {\frac{{n\left( {n + 1} \right)}}{2} - 60} \)…
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