If `(1+2x+x^(2))^(n) = sum_(r=0)^(2n)a_(r)x^(r)`, then `a_(r) = `

Options

A. (a) `(.^(n)C_(r))^(2)`
B. (b) `.^(n)C_(r)..^(n)C_(r+1)`
C. (c) `.^(2n)C_(r)`
D. (d) `.^(2n)C_(r+1)`

Answer

C

Explanation

`(1+2x+x^(2))^(n)=underset(r=0)overset(2n)suma_(r)x^(r)`
or `[(1+x)^(2)]^(n)=underset(r=0)overset(2n)suma_(r)x^(n)`
or `(1+x)^(2n) = underset(r=0)overset(2n)suma_(r)x^(r)`
`underset(r=0)overset(2n)sum.^(2n)C_(r)x^(r) = underset(r=0)overset(2n)suma_(r)x^(r)`
or `a_(r) = .^(2n)C_(r)`