lim _{n} {rightarrow infty} left( frac{sqrt{n | vedclass

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(A) આપેલ પદાવલિને $\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{\sqrt{n}}{\sqrt{(n+4 r)^{3}}}$ તરીકે લખી શકાય.આ સરવાળાને $\sum_{r=0}^{n-1} \fra

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$\frac{5-\sqrt{5}}{10}$

Vedclass · Answer

(A) આપેલ પદાવલિને $\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{\sqrt{n}}{\sqrt{(n+4 r)^{3}}}$ તરીકે લખી શકાય.આ સરવાળાને $\sum_{r=0}^{n-1} \frac{1}{n} \left( \frac{n \sqrt{n}}{\sqrt{(n+4 r)^{3}}} \right) = \sum_{r=0}^{n-1} \frac{1}{n} \left( \frac{1}{(1+\frac{4 r}{n})^{3 / 2}} \right)$ તરીકે લખી શકાય.આ રીમાન સરવાળો છે જે નિશ્ચિત સંકલન $\int_{0}^{1} \frac{dx}{(1+4x)^{3/2}}$ માં રૂપાંતરિત થાય છે.ધારો કે $z = 1+4x$, તો $dz = 4dx$, તેથી $dx = \frac{dz}{4}$.જ્યારે $x=0, z=1$ અને જ્યારે $x=1, z=5$.સંકલન $\frac{1}{4} \int_{1}^{5} z^{-3/2} dz = \frac{1}{4} \left[ \frac{z^{-1/2}}{-1/2} \right]_{1}^{5} = -\frac{1}{2} \left[ \frac{1}{\sqrt{z}} \right]_{1}^{5}$ થશે.$= -\frac{1}{2} \left( \frac{1}{\sqrt{5}} - 1 \right) = \frac{1}{2} \left( 1 - \frac{1}{\sqrt{5}} \right) = \frac{\sqrt{5}-1}{2\sqrt{5}} = \frac{5-\sqrt{5}}{10}$.

$\lim _{n}$ ${\rightarrow \infty} \left( \frac{\sqrt{n}}{\sqrt{n^{3}}}+\frac{\sqrt{n}}{\sqrt{(n+4)^{3}}}+\frac{\sqrt{n}}{\sqrt{(n+8)^{3}}}+\cdots +\frac{\sqrt{n}}{\sqrt{[n+4(n-1)]^{3}}} \right)$ ની કિંમત શોધો.

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