જો mathop {lim }limits_{n to infty } frac{{{1 | vedclass

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

Question

(A) આપેલ લક્ષ $\mathop {\lim }\limits_{n 	o \infty } \frac{{\sum_{r=1}^n r^a}}{{(n+1)^{a-1} \sum_{r=1}^n (na + r)}} = \frac{1}{60}$ છે.અંશ અને છેદને

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) આપેલ લક્ષ $\mathop {\lim }\limits_{n 	o \infty } \frac{{\sum_{r=1}^n r^a}}{{(n+1)^{a-1} \sum_{r=1}^n (na + r)}} = \frac{1}{60}$ છે.અંશ અને છેદને $n^{a+1}$ વડે ભાગતા:$\mathop {\lim }\limits_{n 	o \infty } \frac{{\frac{1}{n} \sum_{r=1}^n (\frac{r}{n})^a}}{{(\frac{n+1}{n})^{a-1} \cdot \frac{1}{n^2} \sum_{r=1}^n (na + r)}} = \frac{1}{60}$.નિશ્ચિત સંકલનની વ્યાખ્યા $\int_0^1 x^a dx = \frac{1}{a+1}$ અને સમાંતર શ્રેણીના સરવાળાનો ઉપયોગ કરતા:$\frac{\int_0^1 x^a dx}{\lim_{n 	o \infty} (1 + \frac{1}{n})^{a-1} (a + \frac{1}{2}(1 + \frac{1}{n}))} = \frac{1}{60}$.$\frac{\frac{1}{a+1}}{a + \frac{1}{2}} = \frac{1}{60}$.$\frac{2}{(a+1)(2a+1)} = \frac{1}{60} \Rightarrow (a+1)(2a+1) = 120$.$2a^2 + 3a - 119 = 0$.$(2a + 17)(a - 7) = 0$.$a > 0$ હોવાથી,$a = 7$ મળે.

Similar Questions

$\lim _{n \rightarrow \infty} \frac{1}{2^{n}}\left(\frac{1}{\sqrt{1-\frac{1}{2^{n}}}}+\frac{1}{\sqrt{1-\frac{2}{2^{n}}}}+\frac{1}{\sqrt{1-\frac{3}{2^{n}}}}+\ldots+\frac{1}{\sqrt{1-\frac{2^{n}-1}{2^{n}}}}\right)$ ની કિંમત શોધો.

$\lim _{n \rightarrow \infty} \frac{3}{n} \left\{ 4 + \left( 2 + \frac{1}{n} \right)^2 + \left( 2 + \frac{2}{n} \right)^2 + \dots + \left( 3 - \frac{1}{n} \right)^2 \right\}$ ની કિંમત શોધો.

$\lim _{n \rightarrow \infty}\left(\frac{n^{2}}{\left(n^{2}+1\right)(n+1)}+\frac{n^{2}}{\left(n^{2}+4\right)(n+2)}+\frac{n^{2}}{\left(n^{2}+9\right)(n+3)}+\ldots+\frac{n^{2}}{\left(n^{2}+n^{2}\right)(n+n)}\right)$ ની કિંમત શોધો.

$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{(2 j-1)+8 n}{(2 j-1)+4 n}$ નું મૂલ્ય શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module