Question

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

• A. $\frac{487}{\sqrt{2^{1945}}}$
• B. $\frac{1946}{\sqrt{2^{1947}}}$
• C. $\frac{1947}{\sqrt{2^{1947}}}$
• D. $\frac{1948}{\sqrt{2^{1947}}}$

Answer 1

The correct option is A $\frac{487}{\sqrt{2^{1945}}}$

Let each term of the ${\sum }_{n=0}^{1947}\frac{1}{2^n+\sqrt{2^{1947}}}$ be represented by $S_n$ where $n$ is the index in the summation.

So, we have $S_n=\frac{1}{2^n+\sqrt{2^{1947}}}$

Consider $S_0+S_{1947}=\frac{1}{1+\sqrt{2^{1947}}}+\frac{1}{2^{1947}+\sqrt{2^{1947}}}$

$=\frac{1}{1+\sqrt{2^{1947}}}(1+\frac{1}{\sqrt{2^{1947}}})$

$=\frac{1}{1+\sqrt{2^{1947}}}\left(\frac{1+\sqrt{2^{1947}}}{\sqrt{2^{1947}}}\right)$

$=\frac{1}{\sqrt{2^{1947}}}$

Similarly,

$S_1+S_{1946}=\frac{1}{2+\sqrt{2^{1947}}}+\frac{1}{2^{1946}+\sqrt{2^{1947}}}$

$=\frac{1}{2}(\frac{1}{1+\sqrt{2^{1945}}}+\frac{1}{2^{1945}+\sqrt{2^{1945}}})$

$=\frac{1}{2}\left(\frac{1}{1+\sqrt{2^{1945}}}\right)\left(\frac{1+\sqrt{2^{1945}}}{\sqrt{2^{1945}}}\right)$

$=\frac{1}{\sqrt{2^{1947}}}$

and so on.

Hence,

${\sum }_{n=0}^{1947}\frac{1}{2^n+\sqrt{2^{1947}}}=\frac{1948}{2}\frac{1}{\sqrt{2^{1947}}}=\frac{974}{\sqrt{2^{1947}}}=\frac{487}{\sqrt{2^{1945}}}$

This is the required solution.