Question

If ${(1+x+x^2)}^{25}={\sum }_{r=0}^{50}{a}_r{x}^r,$ then ${\sum }_{r=0}^{16}{a}_{3r}$ equals

• A. $3^{25}-1$
• B. $3^{24}$
• C. $\frac{3^{25}-1}{3}$
• D. $\frac{3^{24}+1}{3}$

Answer 1

The correct option is B $3^{24}$

$f\left(x\right)={(1+x+x^2)}^{25}$

${\sum }_{r=0}^{16}{a}_{3r}=\frac{1}{3}\left(f\right(1)+f(\omega )+f({\omega }^2\left)\right)$

$f\left(1\right)=3^{25}$

$f\left(w\right)=(1+w+w^2{)}^{25}=0$

$f\left(w^2\right)=(1+w^2+w^4{)}^{25}$
$=(1+w+w^2{)}^{25}=0$ $[\because w^4=w^3.w$ and $w^3=1]$

Hence, $\frac{1}{3}\left(f\right(1)+f(\omega )+f({\omega }^2\left)\right)=\frac{3^{25}}{3}=3^{24}$