If #(1+x)^n# = #C_0 + C_1x + C_2x^2+cdots+C_nx^n#, then #C_1/C_0 + (2C_2)/C_1 + (3C_3)/C_2 + cdots + (nC_n)/C_(n-1) = #?

If #(1+x)^n# = #C_0 + C_1x + C_2x^2+cdots+C_nx^n#, then #C_1/C_0 + (2C_2)/C_1 + (3C_3)/C_2 + cdots + (nC_n)/C_(n-1) = #?

1 Answer
George C.
Feb 23, 2018

#1/2n(n+1)#

Explanation:

Given:

#(1+x)^n = C_0+C_1x+C_2c^2+...+C_nx^n#

By the binomial theorem:

#C_k = ""^nC_k = (n!)/((n-k)!k!)#

So:

#C_(k+1)/C_k = (n!)/((n-k-1)!(k+1)!) -: (n!)/((n-k)!k!)#

#color(white)(C_(k+1)/C_k) = ((n-k)!k!)/((n-k-1)!(k+1)!)#

#color(white)(C_(k+1)/C_k) = ((n-k)color(red)(cancel(color(black)((n-k-1)!)))color(green)(cancel(color(black)(k!))))/(color(red)(cancel(color(black)((n-k-1)!)))(k+1)color(green)(cancel(color(black)(k!))))#

#color(white)(C_(k+1)/C_k) = (n-k)/(k+1)#

So:

#((k+1)C_(k+1))/C_k = n-k#

And:

#sum_(k=0)^(n-1) ((k+1)C_(k+1))/C_k = sum_(k=0)^(n-1) (n-k) = sum_(k=1)^n k = 1/2n(n+1)#

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