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Question
Show That C0 + 2C1 + 3C2 + 4C3 + ... + (n + 1)Cn = (n + 2)2n−1
Solution
C0 + 2C1 + 3C2 + 4C3 + ... + (n + 1)Cn
= C0 + (C1 + C1) + (C2 + 2C2) + (C3 + 3C3) + ... + (Cn + nCn)
= (C0 + C1 + C2 + C3 + ... + Cn) + (C1 + 2C2 + 3C3 + ... + nCn)
= 2n + [nC1+ 2·nC2 + 3·nC3 + ... + n·nCn]
= `2^"n" + ["n" + 2*("n"("n" - 1))/(2!) + 3*("n"("n" - 1)("n" - 2))/(3!) + ... + "n"*1]`
= `2^"n" + "n"[1 + ("n" - 1) + (("n" - 1)("n" - 2))/(2!) + ... + 1]`
= 2n + n[n–1C0 + (n–1)C1 + (n–1)C2 + ... + (n–1)Cn–1]
= 2n + n·2n–1 = 2·2n–1 + n·2n–1
= (n + 2)2n–1
∴ C0 + 2C1 + 3C2 + 4C3 + ... + (n + 1)Cn = (n + 2)2n−1
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