Let k and n be positive integers and put S(k) = 1^k + 2^k + 3^k + ... + n^k.

Show that (m + 1)C(1)S(1) + (m + 1)C(2)S(2) + ... + (m + 1)C(m)S(m) = (n + 1)^(m + 1) - (n + 1)

NOTE: (m + 1)C(1) is (m + 1)choose1

S(k) - k is subscript

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- April 23rd 2009, 07:16 AMfardeen_genBinomial coefficients?
Let k and n be positive integers and put S(k) = 1^k + 2^k + 3^k + ... + n^k.

Show that (m + 1)C(1)S(1) + (m + 1)C(2)S(2) + ... + (m + 1)C(m)S(m) = (n + 1)^(m + 1) - (n + 1)

NOTE: (m + 1)C(1) is (m + 1)**choose**1

S(k) - k is subscript - April 23rd 2009, 09:00 AMPaulRS
In this post we we'll take: to shorten the thing.

Note that:

By the Binomial Theorem:

Now reverse the Order of the sums:

Note that for : thus (Nod)