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Question

The coefficient of x^5 in the expansion of (1+x)21+(1+x)22+.....+(1+x)30 is


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    Solution

    31C621C6

    We have (1+x)21+(1+x)22+....(1+x)30

    =(1+x)21[(1+x)101(1+x)1]

    =1x[(1+x)31(1+x)21]

    Coefficient of x5 in the given expansion =

    Coefficient of x5 in 1x[(1+x)31(1+x)21]

    =Coefficient of x6 in [(1+x)31(1+x)21]

    =31C621C6


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