Question

Let $f\left(n\right)=[\sqrt{n}+\frac{1}{2}]$ (where [x] greatest integer less then equal to $x\left)\forall \right(n\epsilon N)$ Then $\sum _{n-1}^{\infty }\frac{2^{f\left(n\right)}+2^{-f\left(n\right)}}{2^n}$ is equal to __________

• A. $3$
• B. $4$
• C. $2$
• D. $1$

Answer 1

The correct option is A $3$

Let $[\sqrt{n}+\frac{1}{2}]=k\Rightarrow k\le \sqrt{n}+\frac{1}{2}<k+1$
$\Rightarrow {(k-\frac{1}{2})}^2\le n\le {(k+\frac{1}{2})}^2\Rightarrow k^2-k+\frac{1}{4}\le n\le k^2+k+\frac{1}{4}$
$\Rightarrow k^2-k+1\le n\le k^2+k$
So $S=\sum _{n=1}^{\infty }\frac{2^k+2^{-k}}{2^n}=\frac{2+2^{-1}}{2}+\frac{2+2^{-1}}{2^2}+\frac{2^2+2^{-2}}{2^3}+...\frac{2^2+2^{-2}}{2^6}+........\infty$
$=\sum _{k=1}^{\infty }\sum _{n=k^2-k+1}^{k^2+k}\frac{2^k+2^{-k}}{2^n}=\sum _{k=1}^{\infty }(2^k+2^{-k})(2^{-2k^2+k}-2^{-k^2-k})$
$=\sum _{k=1}^{\infty }(2^{-k(k-2)}-2^{-k(k+2)})=(2-2^{-3})+(1-2^{-8})+(2^{-3}-2^{-15})+....=3$