Solve the following?

Let #(1+x)^n=C_0+C_1x+C_2x^2+......+C_nx^n#

Then the value of #(1+C_0/C_1)*(1+C_1/C_2).....*(1+C_(n-1)/C_n)# is?

1 Answer
Apr 15, 2018

Answer:

#(n+1)^n/(n!)#

Explanation:

#(1+C_0/C_1)(1+C_1/C_2)...(1+C_(n-1)/C_n)#

The binomial theorem says that the #k^(th)# coefficient in the #n^(th)# order expansion of #(1+x)# is

#C_k=(n!)/(k!(n-k)!)#

This means that

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

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

So

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

So our product looks like

#((n+1)/(n-0))((n+1)/(n-1))((n+1)/(n-2))...((n+1)/(1))=(n+1)^n/(n!)#