JEE Mains · Maths · STD 11 - 12. limits
\(\lim _{n \rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\ldots .+\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\ldots .+n^3\right)-\left(1^2+2^2+\ldots . .+n^2\right)}\) is equal to:
- A \(\frac{2}{3}\)
- B \(\frac{1}{3}\)
- C \(\frac{3}{4}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\( \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1}\left(r^2-r\right)(n-r)}{\sum_{r=1}^n r^3-\sum_{r=1}^n r^2} \) \( \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1}\left(-r^3+r^2(n+1)-n r\right)}{\left(\frac{n(n+1)}{2}\right)^2-\frac{n(n+1)(2 n+1)}{6}} \)…
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