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About LinkBacksLet 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
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
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
![\sum\limits_{s = 1}^{k-1} {\binom{k}{s}\cdot\left[ {\sum\limits_{j = 0}^{n}{ j^s }} \right]} = \left( {n + 1} \right)^k - (n+1)<br />](http://latex.codecogs.com/png.latex?\sum\limits_{s = 1}^{k-1} {\binom{k}{s}\cdot\left[ {\sum\limits_{j = 0}^{n}{ j^s }} \right]} = \left( {n + 1} \right)^k - (n+1)<br />
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