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Question Number 40216 by scientist last updated on 17/Jul/18

Answered by math1967 last updated on 17/Jul/18

^n c_r + ^n c_(r−1) +^n c_(r−1) + ^n c_(r−2) c_r ^(n+1) + ^(n+1) c_(r−1) = ^(n+2) c_r

ncr+ncr−1+ncr−1+ncr−2cn+1r+n+1cr−1=n+2cr

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