If ` C_(0) , C_(1), C_(2), …, C_(n)` are the binomial coefficients
in the expansion of ` (1 + x)^(n)` , prove that
`(C_(0) + 2C_(1) + C_(2) )(C_(1) + 2C_(2) + C_(3))…(C_(n-1) + 2C_(n) + C_(n+1))`=
`((n+2)^(n))/((n+1)!) prod _(r=1)^(n) (C_(r-1) + C_(r))`.

Explanation

`LHS = (C_(0) + 2C_(1) + C_(2) )(C_(1) + 2C_(2) + C_(3))…(C_(n-1) + 2C_(n) + C_(n+1))`
` prod _(r=1)^(n) (C_(r-1) +2 ""^(n)C_(r) + C_(r+1))`
` prod _(r=1)^(n) {(""^(n)C_(r-1) +""^(n)C_(r))+ (""^(n)C_(r)+ ""^(n)C_(r-1))}`
` prod _(r=1)^(n) (""^(n+1)C_(r-1) + ""^(n+1)C_(r+1))" "` [by Pascal's rule]
` prod _(r=1)^(n) (""^(n+2)C_(r-1) )= prod_(r=1)^(n) ((n+2)/(r+1)) ""^(n+1)C_(r)[because ""^(n)C_(r)= n/r*""^(n-1)C_(r-1)]`
` prod _(r=1)^(n) ((n+2)/(r+1))(""^(n+2)C_(r-1) )= prod_(r=1)^(n) ((n+2)/(r+1))prod_(r=1)^(n) (C_(r-1)+ C_(r))`
` = ((n+2))/(2) *((n+2))/(3) *((n+2))/(4) ...((n+2))/((n+1)) prod_(r=1)^(n) (C_(r-1) + C_(r))`
`((n+2)^(n))/((n+1)!) prod_(r=1)^(n) (C_(r-1) + C_(r))= RHS `