Let S_n be the sum of all integers k such tha | vedclass

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(C) The integers $k$ are $2^n+1, 2^n+2, \dots, 2^{n+1}-1$. This is an arithmetic progression with $a = 2^n+1$,$l = 2^{n+1}-1$,and the number of terms

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$n$ is even

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(C) The integers $k$ are $2^n+1, 2^n+2, \dots, 2^{n+1}-1$. This is an arithmetic progression with $a = 2^n+1$,$l = 2^{n+1}-1$,and the number of terms $N = (2^{n+1}-1) - (2^n+1) + 1 = 2^{n+1}-2^n-1 = 2^n-1$. The sum $S_n$ is given by $S_n = \frac{N}{2}(a+l) = \frac{2^n-1}{2}(2^n+1+2^{n+1}-1) = \frac{2^n-1}{2}(2^n+2 \cdot 2^n) = \frac{2^n-1}{2}(3 \cdot 2^n) = 3 \cdot 2^{n-1}(2^n-1)$. For $9$ to divide $S_n$,we need $3 \cdot 2^{n-1}(2^n-1) = 9m$,which implies $2^{n-1}(2^n-1) = 3m$. This means $2^{n-1}(2^n-1)$ must be a multiple of $3$. Since $2 \equiv -1 \pmod{3}$,we have $2^n \equiv (-1)^n \pmod{3}$. If $n$ is odd,$2^n \equiv -1 \equiv 2 \pmod{3}$,so $2^n-1 \equiv 1 \pmod{3}$. If $n$ is even,$2^n \equiv 1 \pmod{3}$,so $2^n-1 \equiv 0 \pmod{3}$. Thus,$2^n-1$ is divisible by $3$ if and only if $n$ is even.

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

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