The coefficient of x5 in the expansion of (1+x)21+(1+x)22+...+(1+x)30 is
A
51C5
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B
9C5
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C
31C6−21C6
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D
30C5−20C5
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Solution
The correct option is C31C6−21C6 (1+x)21+(1+x)22+...+(1+x)30 =(1+x)21[(1+x)10−1(1+x)−1]=1x[(1+x)31−(1+x)21] ⇒ Coefficient of x5 in the given expression is equal to coefficient of x6 in [(1+x)31−(1+x)21], which is given by 31C6−21C6