mathop {Lim}limits_{n to infty } frac{{{1^2}n | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) ધારો કે અંશ $S_n = \sum_{k=1}^{n} k^2(n - k + 1)$ છે.$S_n = (n+1) \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k^3$.આપણે જાણીએ છીએ કે $\sum_{k=1}^{n} k^2 =

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(A) ધારો કે અંશ $S_n = \sum_{k=1}^{n} k^2(n - k + 1)$ છે.$S_n = (n+1) \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k^3$.આપણે જાણીએ છીએ કે $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ અને $\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}$.છેદ $D_n = \sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}$ છે.તેથી,લક્ષ $\lim_{n 	o \infty} \frac{(n+1) \frac{n(n+1)(2n+1)}{6} - \frac{n^2(n+1)^2}{4}}{\frac{n^2(n+1)^2}{4}}$ થાય.$\frac{n^2(n+1)^2}{4}$ વડે ભાગતા:$= \lim_{n 	o \infty} \left( \frac{(n+1) n(n+1)(2n+1)}{6} \cdot \frac{4}{n^2(n+1)^2} - 1 ight)$.$= \lim_{n 	o \infty} \left( \frac{4n(n+1)^2(2n+1)}{6n^2(n+1)^2} - 1 ight)$.$= \lim_{n 	o \infty} \left( \frac{4(2n+1)}{6n} - 1 ight) = \frac{8}{6} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.

$\mathop {Lim}\limits_{n \to \infty } \frac{{{1^2}n + {2^2}(n - 1) + {3^2}(n - 2) + \dots + {n^2} \cdot 1}}{{{1^3} + {2^3} + {3^3} + \dots + {n^3}}}$ ની કિંમત શોધો :

Similar Questions

$\lim _{x \rightarrow \infty}\left[\frac{8 x^2+5 x+3}{2 x^2-7 x-5}\right]^{\frac{4 x+3}{8 x-1}} = $

$\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos x^2}}{1-\cos x} = $

$\lim _{x \rightarrow 0} \frac{15^{x}-5^{x}-3^{x}+1}{1-\cos 2 x}$ ની કિંમત શોધો.

$\lim _{x \rightarrow 0} x^3 \left\{ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right\} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module