Question

If $(1+x)(1+x+x^2).....(1+x+.....+x^n)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+.....$, then the value of $a_0+a_2+a_4+.....$ is

• A. $\frac{2^n-1}{2}$
• B. $\frac{2^n+1}{2}$
• C. $\frac{(n-1)!}{2}$
• D. $\frac{(n+1)!}{2}$

Answer 1

The correct option is D $\frac{(n+1)!}{2}$

$(1+x)(1+x+x^2).....(1+x+.....+x^n)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+.....$
Put, $x=1,$
$2\times 3\times 4\times ....(n+1)=a_0+a_1+a_2+a_3+......$
$0=a_0-a_1+a_2-a_3+.....$
Put $x=-1,$
$(n+1)!=2[a_0+a_2+....]\Rightarrow a_0+a_2+....=\frac{(n+1)!}{2}$