Question

The value of ${\sum }_{r=0}^{n-1}{a}_{2r}$, is

• A. $\frac{9^n-2a_{2n}-1}{4}$
• B. $\frac{9^n+2a_{2n}+1}{4}$
• C. $\frac{9^n-2a_{2n}+1}{4}$
• D. $\frac{9^n+2a_{2n}-1}{4}$

Answer 1

The correct option is C $\frac{9^n-2a_{2n}+1}{4}$

$(1+x+x^2{)}^{2n}=a_0+a_{1}x+a_2{x}^2...a_{4n}{x}^{4n}$
We know that $a_{4n-r}=a_r$
For $x=1$, we get
$3^{2n}=a_0+a_1+a_2...a_{4n}$ ...(i)
For $x=-1$, we get
$1=a_0-a_1+a_2...a_{4n}$ ...(ii)
Adding i and ii we get
$9^n+1=2[a_0+a_2+a_4+a_6..a_{4n}]$
Applying $a_r=a_{4n-r}$ we get
$\frac{9^n+1}{2}=2[a_0+a_2+a_4+a_6...a_{2n-r}]+a_{2n}$
$\frac{9^n+1-2a_{2n}}{4}=a_0+a_2+a_4+a_6...a_{2n-r}$