The expressions `1+x,1+x+x^2,1+x+x^2+x^3.............1+x+x^2+..............+x^a` are mutiplied together and the terms of the product thus obtained are arranged in increasing powers of `x` in the from of `a_0+a_1x+a_2x^2+.................,` then sum of even coefficients?

Options

A. `20!`
B. `21!`
C. `(21!)/(2)`
D. `19!`

Answer

C

Explanation

`(c )` Let `f(x)=(1+x)(1+x+x^(2))(1+x+x^(2)+x^(3))…(1+x+x^(2)+….+x^(20))`
Highest degree of `f(x)`
`=` highest degree of `x` present in the `f(x)`
`=1+2+3+…+20`
`=(20(21))/(2)=210`
Since in the expansion of `f(x)` degree of `x` from zero to `210`, all present so all terms containing `x^(0)`, `x^(1)`, `x^(2)`.....`x^(210)` will be present
`:.` total no. of terms `=210+1=211`
`(1+x)(1+x+x^(2))....(1+x+x^(2)+....+x^(20))`
`=a_(0)+a_(1)x+a_(2)x^(2)+....+a_(210)x^(210)`......`(i)`
Replacing `x` by `1//x`, we get
`(1+(1)/(x))(1+(1)/(x)+(1)/(x^(2)))....(1+(1)/(x)+(1)/(x^(2))+....+(1)/(x^(20)))`
`=a_(0)+(a_(1))/(x)+(a_(2))/(x^(2))+....+(a_(210))/(x_(210))`
Taking `L.C.M` both sides, we get
`(x+1)(x^(2)+x+1)....(x^(20)+x^(19)+....+x+1)`
`=a_(0)x^(210)+a_(1)x^(209)+...+a_(209)x+a_(210)`
Thus `a_(0)+x^(210)+a_(1)x^(209)+....+a_(209)x+a_(210)`
Thus `a_(r )=a_(210-r)`
`a_(0)+a_(1)x+a_(2)x^(2)+a_(3)x^(3)+.....`
`=(1+x)(1+x+x^(2))...(1+x+x^(2)+...+x^(20))`
Putting `x=1` in `(i)`
`:. a_(0)+a_(1)+a_(2)+....=21!`......`(ii)`
Again putting `x=-1`
`a_(0)-a_(1)+a_(2)-a_(3)+a_(4)=0`.........`(iii)`
Adding equation `(ii)` and `(iii)`, we have
`2[a_(0)+a_(2)+a_(4)+....]=21!`
`:.a_(0)+a_(2)+a_(4)+....=(21!)/(2)=` sum of even coefficients