JEE Mains · Maths · STD 11 - 8. sequence and series
\(1\, + \,\frac{{{1^3}\, + \,{2^3}}}{{1 + 2}} + \frac{{{1^3}\, + \,{2^3} + {3^3}}}{{1 + 2 + 3}} + ...... + \frac{{{1^3}\, + \,{2^3} + {3^3} + ..... + {{15}^3}}}{{1 + 2 + 3 + ..... + 15}} - \frac{1}{2}\left( {1 + 2 + 3 + ....+15} \right)\) is equal to
- A \(620\)
- B \(1860\)
- C \(1240\)
- D \(660\)
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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