Question

Let $S_n$ be the sum of all integers k such that $2^n<k<2^{n-1}$, for n > 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

Number of integers between 2${}^n$ and $2^{2n+1}-2^n-11$
and I term $=2^n+1$
last term $=2^{n+1}-1$
$\therefore S_n=\frac{(2^{n+1}-2^n-1)}{2}[2^n+1+2^{n+1}-1]$
$\frac{(2^{n+1}-2^n-1)}{2}\left(2^n\right)(1+2)$
$=(2^n-1)\frac{\left(2^n\right).3}{2}$
$S_n=9\lambda ;\lambda \epsilon I$
$\therefore 3\times 2^{n-1}\times (2^n-1)=9\lambda$
$2^{n-1}\times (2^n-1)=3\lambda$
$2^n(2^n-1)=6\lambda$
It is possible when n is even.