Question

If $nϵN$ and $(1+4x+4x^2{)}^n=\sum _{r=0}^{r=2n}{a}_r{x}^r$ then value of $2\sum _{r=0}^n{a}_{2r}$ equals

• A. $9^n-1$
• B. $9^n+1$
• C. $3^n+1$
• D. $3^n-1$

Answer 1

The correct option is B $9^n+1$

$\because 1+4x+4x^2=(1+2x{)}^2$
$\therefore (1+2x{)}^{2n}=\sum _{r=0}^{2n}{a}_r{x}^r$
putting $x=1$
$3^{2n}=\sum _{r=0}^{2n}{a}_r=a_0+a_1+a_2+a_3+.....+a_{2n}....\left(i\right)$
and putting $x=-1$
$1=\sum _{r=0}^{r=2n}{a}_r(-1{)}^r=a_0-a_1+a_2-a_3+....+a_{2n}...(ii)$
Now adding (i) and (ii) we get
$2(a_0+a_2+a_4+....+a_{2n})=3^{2n}+1$
$\Rightarrow 2\sum _{r=0}^{r=n}{a}_{2r}=9^n+1$
Hence choice (b) is correct answer.