The coefficient of x5 in the expansion of (1+x)21+(1+x)22+(1+x)23+⋯+(1+x)30
A
31C6−21C6
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B
31C6+21C6
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C
2.31C6+3.21C6
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D
31C6+10⋅21C6
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Solution
The correct option is A31C6−21C6 (1+x)21+(1+x)22+(1+x)23+⋯+(1+x)30=(1+x)21{(1+x)10−1}(1+x)−1=1x[(1+x)31−(1+x)21] coefficient of x5 = coefficient of x6 in [(1+x)31−(1+x)21] coefficient of x5=31C6−21C6