Question

If ${(1+x+x^2)}^{25}={\sum }_{r=0}^{50}{a}_r{x}^r,$ then ${\sum }_{r=0}^{12}{a}_{4r}$ equals

• A. $\frac{3^{25}+1}{4}$
• B. $\frac{3^{25}-1}{4}$
• C. $\frac{3^{24}+1}{2}$
• D. $\frac{3^{25}-1}{2}$

Answer 1

The correct option is A $\frac{3^{25}+1}{4}$

${\sum }_{r=0}^{12}{a}_{4r}=\frac{1}{4}\left(f\right(1)+f(-1)+f(i)+f(-i\left)\right)$

$=\frac{1}{4}(3^{25}+1+i-i)$

$=\frac{1}{4}(3^{25}+1)$