1 + left( frac{1}{2} + frac{1}{3} right) frac | vedclass

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(A) Let the given series be $S = 1 + \sum_{n=1}^{\infty} \left( \frac{1}{2n} + \frac{1}{2n+1} \right) \frac{1}{4^n}$.$S = 1 + \sum_{n=1}^{\infty} \fra

Vedclass · Accepted Answer

$\log_e (2\sqrt{3})$

Vedclass · Answer

(A) Let the given series be $S = 1 + \sum_{n=1}^{\infty} \left( \frac{1}{2n} + \frac{1}{2n+1} \right) \frac{1}{4^n}$.$S = 1 + \sum_{n=1}^{\infty} \frac{1}{2n} \left( \frac{1}{4} \right)^n + \sum_{n=1}^{\infty} \frac{1}{2n+1} \left( \frac{1}{4} \right)^n$.Using the expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$ and $\log_e \left( \frac{1+x}{1-x} \right) = 2 \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \dots \right)$.Let $x = \frac{1}{2}$. Then $\log_e(1+x) = \log_e(3/2)$ and $\log_e \left( \frac{1+x}{1-x} \right) = \log_e(3)$.Substituting these values,we get $S = \log_e(2\sqrt{3})$.

$1 + \left( \frac{1}{2} + \frac{1}{3} \right) \frac{1}{4} + \left( \frac{1}{4} + \frac{1}{5} \right) \frac{1}{4^2} + \left( \frac{1}{6} + \frac{1}{7} \right) \frac{1}{4^3} + \dots \infty = $

Similar Questions

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