MCQ
Let $S_n$ be the sum of all integers $k$ such that $2^n < k < 2^{n+1}$, for $n \geq 1$. Then,$9$ divides $S_n$ if and only if
- A$n$ is odd
- B$n$ is of the form $3 k+1$
- ✓$n$ is even
- D$n$ is of the form $3 k+2$
We have, $2^n < k < 2^{n+1}, k \in N$
Number of integer between $2^n$ and $2^{n+1}$ is i.e $k=2^{n+1}-2^n-1$
First term $=2^n+1$
Last term $=2^{n+1}-1$
$\therefore S_n=\frac{2^{n+1}-2^n-1}{2}\left[2^n+1+2^{n+1}-1\right]$
$S_n=\frac{2^{n+1}-2^n-1}{2}\left(2^n\right)(1+2)$
$S_n=\frac{\left(2^n-1\right)\left(2^n\right) \cdot 3}{2}$
But $S_n=9 m, m \in I$
$\therefore \quad \frac{\left(2^n-1\right) 2^n \cdot 3}{2}=9 m$
$\Rightarrow \quad\left(2^n-1\right) 2^{n-1}=3\,m$
$\Rightarrow \quad 2^n\left(2^n-1\right)=6\,m$
It is possible when, $n$ is even.
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