Question #d7964

1 Answer
George C.
Dec 4, 2017

The coefficient of #x^k# is:

#((n+1),(k+1)) = ((n+1)!)/((n-k)! (k+1)!)#

Explanation:

Note that:

#(t-1)(t^n+t^(n-1)+...+t+1) = t^(n+1)-1#

So putting #t=(1+x)#, we find:

#xE = ((1+x)-1)E#

#color(white)(xE) = ((1+x)-1)((1+x)^n+(1+x)^(n-1)+...+(1+x)+1)#

#color(white)(xE) = (1+x)^(n+1) - 1#

#color(white)(xE) = (sum_(k=0)^(n+1) ((n+1),(k)) 1^(n+1-k) x^k) - 1#

#color(white)(xE) = sum_(k=1)^(n+1) ((n+1),(k)) x^k#

So:

#E = sum_(k=1)^(n+1) ((n+1),(k)) x^(k-1) = sum_(k=0)^n ((n+1),(k+1)) x^k#

The coefficient of #x^k# is:

#((n+1),(k+1)) = ((n+1)!)/((n-k)! (k+1)!)#

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