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

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

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