If `(4x^(2) + 1)^(n) = sum_(r=0)^(n)a_(r)(1+x^(2))^(n-r)x^(2r)`, then the value of `sum_(r=0)^(n)a_(r)` is

Options

A. `3^(n)`
B. `4^(n)`
C. `5^(n)`
D. `6^(n)`

Answer

B

Explanation

`(4x^(2)+1)^(n)=underset(r=0)overset(n)suma_(r)(1+x^(2))^(n-r)x^(2r)`
`rArr (1+(3x^(2))/(1+x^(2)))^(n)=underset(r=0)overset(n)suma_(r)((x^(2))/(1+x^(2)))^(r)`
`:. a_(r) = .^(n)C_(r)3^(r)`
`:. underset(r=0)overset(n)suma_(r) = underset(r=0)overset(n)sum.^(n)C_(r)3^(r)=(1+3)^(n)=4^(n)`