If ` (1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + …+ C_(n) x^(n)` , prove that
` C_(0) *""^(2n)C_(n) - C_(1) *""^(2n-2)C_(n) + C_(2) *""^(2n-4) C_(n) -…= 2^(n)`

Answer

A, B

Explanation

`(a,b)` The given series can be written as
`S=sum_(r=0)^(n)'^(n)C_(r )^(2n-2)C_(m)(-1)^(r )`
`=sum_(r=0)^(n)'^(n)C_(r )(-1)^(r )xx"coefficient of" x^(m) "in" (1+x)^(2n-2r)`
`="coefficient of" x^(m) "in" sum_(r=0^(n)'^(n)C_(r )(-1)^(r )[(1+x)^(2)]^(n-r)`
`="coefficient of" x^(m) "in" (x^(2)+2x)^(n)`
`="coefficient of" x^(m) "in" x^(n) (x+2)^(n)`
`="coefficient of" x^(m-n) "in" (x+2)^(n)`
`=^(n)C_(m-n)2^(n-(m-n))=((n),(m-n))2^(2n-m)` if `m ge n` and `0` if `m lt n`.