`" if" C_(0)C_(1)C_(2),……C_(n)` are the binomial coefficients in the expansion of `(1+x)^(n)` then prove that:
`C_(0)C_(2)+C_(1)C_(3)+C_(2)C_(4)+……+C_(n-2)C_(n)=(|ul2n)/(|uln-2|uln+2)`

Answer

N/a

Explanation

`(1+x)^(n)=c_(0)+C_(1).x+C_(2).x^(2)+C_(3).x^(3)+......+C_(n).x^(n)`
`" and " (x+1)^(n)=C_(0).x^(n)+C_(1).x^(n)+C_(1).x^(n-1)+C_(2).x^(n-2)`
`+C_(3).x^(n-3)+…..+C_(n)`
`" Now "(1+x)^2n)=(C_(0)+C_(1).x+C_(2).x^(2)+…+C_(n).x^(n))`
`(C_(0).x^(n)+C_(1).x^(n-1)+C_(2).x^(n-2)`
`+C_(3).x^(n-3).....+C_(n))`
`" Now " C_(0)C_(2)+C_(1)C_(3)+C_(2)C_(4)+.....+C_(n-2)C_(n)=" Coefficient of "x^(n-2)` in the expansion of `(1+x)^(2n)`
`:.` In the expansion of `(1+x)^(2n)`
`T_(r+1)=^(2n)C_(r).(1)^(2n-r).x^(r)`
`=^(2n)C_(r).x^(r)`
`:.` Coefficient of `x^(n-2)=^(2n)C_(n-2)=(|ul2n)/(|ul(n-2)|ul(n+2))`
Therefore
`C_(0)C_(2)+C_(1)C_(3)+C_(2)C_(4)+......+C_(n-2)C_(n)=(|ul2n)/(|ul(n-2)|uln+2)`
Hence Proved.