Define a function of x and n as follows:
F = Sum(j=1 to n) of xj+1 nCj (1)
= x[Sum(j=1 to n) of xj nCj ]
=x[[Sum(j=0 to n) of xj nCj ] - 1]
= x[(1 + x)n - 1] (2)
Using D(y) to denote the derivative of y wrt x,
(1) gives D(D(F)) = Sum(j=1 to n) of j(j + 1)xj-1 nCj
so x*D(D(F)) = Sum(j=1 to n) of j(j + 1)xj nCj
so D(x*D(D(F))) = Sum(j=1 to n) of j2(j + 1)xj-1 nCj
Using (2) and switching sides now gives:
Sum(j=1 to n) of j2(j + 1)xj-1 nCj
= D(x*D(D(x[(1 + x)n - 1])))
= D(x*D((1 + x)n + nx(1 + x)n-1 - 1))
= D(x*(n(1 + x)n-1 + n(n - 1)x(1 + x)n-2 + n(1 + x)n-1))
= D(nx(1 + x)n-1 + n(n - 1)x2(1 + x)n-2 + nx(1 + x)n-1)
= D(2nx(1 + x)n-1 +n(n - 1)x2(1 + x)n-2
= 2n(n-1)x(1+x)n-2 + 2n(1+x)n-1 + n(n-1)[(n-2)x2(1+x)n-3 + 2x(1+x)n-2]
Now substituting x = 1 into this equation gives:
Sum(j=1 to n) of j2(j + 1) nCj
= 2n-22n(n - 1) + 2n-12n + n(n - 1)[2n-3(n - 2) + 2n-22]
= 2n-3n[4(n - 1) + 8 + (n - 1)(n + 2)]
= 2n-3n(n2 + 5n + 2)
which agrees with Broll's result and with Charlie's tabulations.
Edited on April 29, 2011, 11:36 pm
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Posted by Harry
on 2011-04-29 23:34:28 |
Solution
