Question

Let $(1-x-2x^2{)}^6=1+a_{1}x+a_2{x}^2+....+a_{12}{x}^{12}$. Then $\frac{a_2}{2^2}+\frac{a_4}{2^4}+\frac{a_6}{2^6}+....+\frac{a_{12}}{2^{12}}$ is equal to?

• A. $-1$
• B. $-1/2$
• C. $0$
• D. $1/2$

Answer 1

The correct option is A $-1$

Putting $x=\frac{1}{2}$,we get

$(1-\frac{1}{2}-\frac{1}{2}{)}^6=1+\frac{a_1}{2}+\frac{a_2}{2^2}+.......+\frac{a_{12}}{2^{12}}$

Putting $x=-\frac{1}{2}$ , we get

$(1+\frac{1}{2}-\frac{1}{2}{)}^6=1-\frac{a_1}{2}+\frac{a_2}{2^2}-\frac{a_3}{2^3}+.....+\frac{a_{12}}{12}$

Adding these equations, we get

$1=2+\frac{a_2}{2^2}+\frac{a_4}{2^4}+....+\frac{a_{12}}{2^{12}}$

$-1=\frac{a_2}{2^2}+\frac{a_4}{2^4}+....+\frac{a_{12}}{2^{12}}$