The sum of the series 101C1 101C0+2⋅101C2 101C1+3⋅101C3 101C2+…+101.101C101 101C100 is __________.

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answer is 5151.

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Detailed Solution

Here, Tr=r⋅101Cr 101Cr−1=r⋅(101−r+1)r=102−r∴ Given sum =∑r=1101 (102−r) =101+100+99+…+2+1 =5151

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