If `(1+x^(2))^(n) = underset(r=0)overset(n)suma_(r)x^(r )= (1+x+x^(2)+x^(3))^(100)`. If `a = underset(r=0)overset(300)suma_(r)`, then `underset(r=0)overset(300)sumra_(r)` is

Options

A. 300a
B. 100a
C. 150a
D. 75a\

Answer

C

Explanation

`underset(r=0)overset(300)suma_(r) xx x^(r)=(1+x+x^(2)+x^(3))^(100)`
Clearly. `'a_(r)'` is the coefficient of `x^(r)` in the expansion of `(1+x+x^(2)+x^(3))^(100)`.
Replacing x by `1//x` in the given equation, we get
`underset(r=0)overset(300)suma_(r)(1/x)^(r)=(1)/(x^(300))(x^(3)+x^(2)+x+1)^(100)`
or `underset(r=0)overset(300)suma_(r)x^(300-r)=(1+x+x^(2)+x^(3))^(100)`
Here, `a_(r)` represents the coefficient of `x^(300-r)` in `(1+x+x^(2)+x^(3))^(100)`.
Thus, `a_(r) = a_(300-r)`
Let `I=underset(r=0)overset(300)sumrxxa_(r)`
`=underset(r=0)overset(300)sum(300-r)a_(300-r)`
`=underset(r=0)overset(300)sum(300-r)a_(r)`
`=300underset(r=0)overset(300)suma_(r)-underset(r=0)overset(300)sumra_(r)`
`rArr 2I = 300a`
or `I = 150a`