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Question

(A) આપેલ છે $y = \sum_{n=1}^{\infty} \frac{n}{n+1} x^{n+1} = \sum_{n=1}^{\infty} (1 - \frac{1}{n+1}) x^{n+1}$.$y = \sum_{n=1}^{\infty} x^{n+1} - \sum_

Vedclass · Accepted Answer

$\frac{1}{2} e^{2}$

Vedclass · Answer

(A) આપેલ છે $y = \sum_{n=1}^{\infty} \frac{n}{n+1} x^{n+1} = \sum_{n=1}^{\infty} (1 - \frac{1}{n+1}) x^{n+1}$.$y = \sum_{n=1}^{\infty} x^{n+1} - \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}$.$y = (x^2 + x^3 + x^4 + \dots) - (\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots)$.અનંત ગુણોત્તર શ્રેણીના સરવાળા $\frac{x^2}{1-x}$ અને વિસ્તરણ $-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$ નો ઉપયોગ કરતા:$y = \frac{x^2}{1-x} - (-\ln(1-x) - x) = \frac{x^2}{1-x} + \ln(1-x) + x$.$y = \frac{x^2 + x(1-x)}{1-x} + \ln(1-x) = \frac{x}{1-x} + \ln(1-x)$.$x = \frac{1}{2}$ માટે,$y = \frac{1/2}{1-1/2} + \ln(1-1/2) = 1 + \ln(1/2) = 1 - \ln 2$.તેથી $e^{1+y} = e^{1 + 1 - \ln 2} = e^{2 - \ln 2} = e^2 \cdot e^{-\ln 2} = e^2 \cdot \frac{1}{2} = \frac{e^2}{2}$.

જો $0 < x < 1$ અને $y = \frac{1}{2} x^{2} + \frac{2}{3} x^{3} + \frac{3}{4} x^{4} + \dots$ હોય,તો $x = \frac{1}{2}$ આગળ $e^{1+y}$ ની કિંમત શું થાય?

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