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Question

Let sn be the sum of all integers k such that 2n<k<2n+1, for n1. Then 9 divides Sn, if and only if

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

The correct option is C n is even
sn=(2n+1)+(2n+2)+(2n+3)+.......+(2n+2n1)
sn=(2n1)×(2n)+(1+2+3......+(2n1))
sn=(2n1)×(2n)+(2n1)×(2n)2
sn=(2n1)×3×2n1
For it to be divisible by 9, (2n1) must bedivisible by 3.
Therefore n must be even.

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