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

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(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

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$7$

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(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

$\mathop {\lim }\limits_{n \to \infty } {\left\{ {\left( {1 + \frac{{{1^2}}}{{{n^2}}}} \right)\left( {1 + \frac{{{2^2}}}{{{n^2}}}} \right)\left( {1 + \frac{{{3^2}}}{{{n^2}}}} \right) \dots \left( {1 + \frac{{{{(n - 1)}^2}}}{{{n^2}}}} \right)} \right\}^{1/n}}$ ની કિંમત શોધો:

જો $\lim _{n}$ ${\rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right)\left(1+\frac{9}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{\frac{1}{n}}=ae^{b}$ હોય,તો $a+b=$

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

જો $a = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{2n}{n^2+k^2}$ અને $f(x) = \sqrt{\frac{1-\cos x}{1+\cos x}}$,$x \in (0, 1)$,હોય તો:

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