Q56 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics
If the sum of the coefficients of even powers of $x$ in the expansion of $\left(1-x+x^2\right)^{2 n}$ is 3281 , then $\mathrm{n}=$
- Correct answerA4
- B5
- C6
- D3
Solution
Let $\left(1-x+x^2\right)^{2 n}=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{4 n}$ put $x=1$, we get
$$
1^{2 \mathrm{n}}=1=\mathrm{a}_0+\mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{2 \mathrm{n}}.....(i)
$$
Put $x=-1$, we get
$$
3^{2 n}=a_0-a_1+a_2-\ldots+a_{2 n}.....(ii)
$$
Adding equations (i) and (ii) we get
$$
1+3^{2 n}=2\left(a_0+a_2+\ldots+a_{2 n}\right)
$$
Given that sum of even power of $x$ is 3281
$$
\begin{aligned}
& \Rightarrow 1+3^{2 n}=2 \times 3281=6562 \\
& \Rightarrow 3^{2 n}=6561=3^8 \Rightarrow 2 n=8 \Rightarrow n=4
\end{aligned}
$$
Hence, the correct answer is option: 'A'.