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

$T\left(n\right)={\sum }_{n=0}^{1947}\frac{1}{2^n+\sqrt{2^{1947}}}$

We can see,

$T\left(0\right)=\frac{1}{2^0+\sqrt{2^{1947}}}$

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

And,

$T\left(0\right)+T\left(1947\right)=\frac{1}{\sqrt{2^{1947}}}$

Similarly,

$T\left(1\right)+T\left(1946\right)=\frac{1}{\sqrt{2^{1947}}}$

$and$ $so$ $on.$

We can see

There are $1948$ terms and $974$ pairs

$\therefore$

$Sum$ $is$ $\frac{974}{\sqrt{2^{1947}}}$

$\Rightarrow$ $\frac{974}{\sqrt{2^2\times 2^{1945}}}=\frac{487}{\sqrt{2^{1945}}}$