Q105 / 160 · TS EAMCET 2023 (12 May Shift 1) · Mathematics
The coefficient of $x^{50}$ in the expansion of $(1+x)^{101}\left(1-x+x^2\right)^{100}$ is
- Correct answerA$0$
- B$-1$
- C$50$
- D$100$
Solution
$\begin{aligned}
& (1+x)^{101}\left(1-x+x^2\right)^{100} \\
& =(1+x)\left[(1+x)\left(1-x+x^2\right)\right]^{100} \\
& =(1+x)\left(1+x^3\right)^{100}=1\left(1+x^3\right)^{100}+x\left(1+x^3\right)^{100} \\
& \begin{aligned}
\therefore \text { Coefficient of } x^{50} \\
=\left\{1 \times \text { coefficient of } x^{50} \text { in }\left(1+x^3\right)^{100}\right\} \\
\quad+\left\{1 \times \text { coefficient of } x^{49} \text { in }\left(1+x^3\right)^{100}\right\}
\end{aligned}
\end{aligned}$
$\because\left(1+x^3\right)_n$ will contain only powers of $x$ which are multiple of 3 .
$\therefore \quad$ Coefficient of $x^{50}=$ Coefficient of $x^{49}=0$
Hence, the correct answer is option: 'A'.