Question

If n is a positive integer and ${(1+x+x^2)}^n=\sum _{r=0}^{2n}{a}_r{x}^r$ ,then the value $a_0+a_1+a_2+a_3\cdots a_{n-1}$ equals

• A. $\frac{3^n+a_n}{2}$
• B. $\frac{3^n+a_{n+r}}{2}$
• C. $\frac{3^n-a_n}{2}$
• D. None of these

Answer 1

The correct option is C $\frac{3^n-a_n}{2}$

$(1+x+x^2{)}^n=a_0+a_1{x}^1+a_2{x}^2+...a_{2n}{x}^{2n}$
Hence
$a_r=a_{2n-r}$
Substituting $x=1$ we get
$3^n=a_0+a_1+a_2+...a_{2n}$
$a_r=a_{2n-r}$
Therefore
$3^n=a_0+a_1+a_2+...a_n+...a_{2n-2}+a_{2n-1}+a_{2n}$
$=(a_0+a_{2n})+(a_1+a_{2n-1})+(a_2+a_{2n-2})+...(a_{n-1}+a_{n+1})+a_n$
$=2[a_0+a_1+a_2+a_3+...a_{n-1}]+a_n$
$\frac{3^n-a_n}{2}=a_0+a_1+a_2+a_3+...a_{n-1}$