(a)

We have,

$\sum \limits_{n=0}^{1947} \frac{1}{2^n+\sqrt{2^{1947}}}$

Let $\quad f(n)=\frac{1}{2^n+\sqrt{2^{1947}}}$

$f(0)+f(1947)=\frac{1}{1+2^{\frac{1947}{2}}}+\frac{1}{2^{1947}+2^{\frac{1947}{2}}}$

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

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

Similarly, $f(1)+f(1946)=\frac{1}{2^{\frac{1947}{2}}}$

$\because \sum \limits_{n=0}^{1947} f(x)=f(0)+f(1)+f(2)+f(3)+\ldots .+f(1947)$

$=(f(0)+f(1947)+(f(1)+f(1946))+\ldots +(f(973)+f(974))$

$=974 \times \frac{1}{2^{\frac{1917}{2}}}$

$\sum \limits_{n=0}^{1947} f(n)=\frac{2 \times 487}{2 \times 2^{\frac{1945}{2}}}=\frac{487}{\sqrt{2^{1945}}}$