JEE Mains · Maths · STD 11 - 8. sequence and series
If \(\frac{1^3+2^3+3^3+\ldots \ldots \text {.upto } n \text { terms }}{1 \cdot 3+2 \cdot 5+3 \cdot 7+\ldots \ldots \text { upto } n \text { terms }}=\frac{9}{5}\), then the value of \(n\) is
- A \(10\)
- B \(15\)
- C \(20\)
- D \(5\)
Answer & Solution
Correct Answer
(D) \(5\)
Step-by-step Solution
Detailed explanation
\(1^3+2^3+3^3 \ldots . .+n^3=\left(\frac{n(n+1)}{2}\right)^2\) \(1 \cdot 3+2 \cdot 5+3 \cdot 7+\ldots \ldots+ n \text { terms }=\) \(\sum \limits_{r=1}^n r(2 r+1)=\sum \limits_{r=1}^n\left(2 r^2+r\right)\) \(=\frac{2 \cdot n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\)…
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