The value of sum limits_{n=0}^{1947} frac{1}{ | vedclass

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Question

(A) Let $S = \sum \limits_{n=0}^{1947} f(n)$ where $f(n) = \frac{1}{2^n + \sqrt{2^{1947}}}$.Note that the number of terms in the sum is $1947 - 0 + 1

Vedclass · Accepted Answer

$\frac{487}{\sqrt{2^{1945}}}$

Vedclass · Answer

(A) Let $S = \sum \limits_{n=0}^{1947} f(n)$ where $f(n) = \frac{1}{2^n + \sqrt{2^{1947}}}$.Note that the number of terms in the sum is $1947 - 0 + 1 = 1948$.Consider the sum of terms $f(n) + f(1947-n)$:$f(n) + f(1947-n) = \frac{1}{2^n + \sqrt{2^{1947}}} + \frac{1}{2^{1947-n} + \sqrt{2^{1947}}}$$= \frac{1}{2^n + \sqrt{2^{1947}}} + \frac{1}{\frac{2^{1947}}{2^n} + \sqrt{2^{1947}}}$$= \frac{1}{2^n + \sqrt{2^{1947}}} + \frac{2^n}{2^{1947} + 2^n \sqrt{2^{1947}}}$$= \frac{1}{2^n + \sqrt{2^{1947}}} + \frac{2^n}{\sqrt{2^{1947}}(\sqrt{2^{1947}} + 2^n)}$$= \frac{\sqrt{2^{1947}} + 2^n}{\sqrt{2^{1947}}(2^n + \sqrt{2^{1947}})} = \frac{1}{\sqrt{2^{1947}}}$.Since there are $1948$ terms,we can pair them into $1948/2 = 974$ pairs.Thus,$S = 974 	imes \frac{1}{\sqrt{2^{1947}}} = \frac{974}{2^{1947/2}} = \frac{2 	imes 487}{2^{1947/2}} = \frac{487}{2^{1947/2 - 1}} = \frac{487}{2^{1945/2}} = \frac{487}{\sqrt{2^{1945}}}$.Therefore,the correct option is $A$.

The value of $\sum \limits_{n=0}^{1947} \frac{1}{2^n+\sqrt{2^{1947}}}$ is equal to

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