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

\(\frac{2^{3}-1^{3}}{1 \times 7}+\frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 \times 11}+\)\(\frac{6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}}{3 \times 15}+\ldots .+\) \(\frac{30^{3}-29^{3}+28^{3}-27^{3}+\ldots+2^{3}-1^{3}}{15 \times 63}\)is equal to.

  1. A \(140\)
  2. B \(130\)
  3. C \(120\)
  4. D \(110\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(120\)

Step-by-step Solution

Detailed explanation

\(T_{n}=\frac{2 \sum_{r=1}^{n}(2 r)^{3}-\left(\sum_{T=1}^{2 n} r^{3}\right)}{n(4 n+3)}\) \(T _{ n }=1\,n\) So, \(\sum_{n=1}^{15} T_{n}=120\)
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