If ∑(r = 0 to n) (nCr)^-1 = s, then prove that ∑(r = 0 to n) (2r + 1)/(nCr) = (n+1)s nCr is n choose r.
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Originally Posted by fardeen_gen If ∑(r = 0 to n) (nCr)^-1 = s, then prove that ∑(r = 0 to n) (2r + 1)/(nCr) = (n+1)s nCr is n choose r. Let and . Notice that, now let and so . Therefore, Thus, .
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