Question

If $c_0,c_1,c_2,.......c_n$ denote the coefficients in the expansion of ${(1+x)}^n$, prove that
$(c_0+c_1)(c_1+c_2).......(c_{n-1}+c_n)=\frac{c_1{c}_2...c_n{(n+1)}^n}{|\underset{–}{n}}$.

Answer 1

$(c_0+c_1)(c_1+c_2)\dots (c_{n-1}+c_n)=\frac{c_1{c}_2\dots c_n{(n+1)}^n}{n!}$

This can be rearranged as,

$\left(\frac{c_0+c_1}{c_1}\right)\left(\frac{c_1+c_2}{c_2}\right)\dots \dots \left(\frac{c_{n-1}+c_n}{c_n}\right)=\frac{(n+1{)}^n}{n!}$

Or $\prod _{r=0}^{n-1}\left(\frac{c_r+c_{r+1}}{c_{r+1}}\right)=\frac{(n+1{)}^n}{n!}$

Also, $\left(\frac{c_r+c_{r+1}}{c_{r+1}}\right)=(\frac{c_r}{c_{r+1}}+1)=\frac{n!}{(n-r-1)!(r+1)!}\times \frac{(n-r)!r!}{n!}+1=\frac{n+1}{n-r}$

Therefore, $L.H.S=\prod _{r=0}^{n-1}\left(\frac{c_r+c_{r+1}}{c_{r+1}}\right)=\prod _{r=0}^{n-1}\left(\frac{n+1}{n-r}\right)=\frac{(n+1{)}^n}{n!}=R.H.S$