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

Answer: (C)

n is even

Sum of all integers between $2^n$ and $2^{n+1}$ is:

$\therefore S=sum upto $2^{n+1}$-sum upto $2^n$

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

$=2^{2n+1}-2^n-2^{2n-1}-2^{n-1}$

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

$=\frac{3}{2}{.2}^n(2^n-1)$

$\therefore (2^n-1)$ is divisible by 3 for $n=even$

$\therefore 3(2^n-1)$ is divisible by 9 for $n=even$