Question

Let $s_n$ be the sum of all integers k such that $2^n<k<2^{n+1},$ for $n\ge 1.$ Then $9$ divides $S_n$, if and only if

• A. $n$ is odd
• B. $n$ is of the form $3k+1$
• C. $n$ is even
• D. $n$ is of the form $3k+2$

Answer 1

The correct option is C $n$ is even

$s_n={(2}^n+1)+{(2}^n+2)+{(2}^n+3)+.......+{(2}^n+2^n-1)$

$s_n=(2^n-1)\times {(2}^n)+(1+2+\mathrm{3......}+{(2}^n-1\left)\right)$

$s_n=(2^n-1)\times {(2}^n)+\frac{(2^n-1)\times {(2}^n)}{2}$

$s_n=(2^n-1)\times 3\times 2^{n-1}$

For it to be divisible by $9$, $(2^n-1)$ must bedivisible by $3$.

Therefore $n$ must be even.