Find the coefficient of $x$ in the expansion of $\left(1-3 x+7 x^{2}\right)(1-x)^{16}$.
To Find : coefficient of x
Formula : $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$
We have a formula,
$\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$
Therefore, expansion of $(1-x)^{16}$ is given by,
$(1-x)^{16}=\sum_{r=0}^{16}\left(\begin{array}{c}16 \\ r\end{array}\right)(1)^{16-r}(-x)^{r}$
$=\left(\begin{array}{c}16 \\ 0\end{array}\right)(1)^{16}(-x)^{0}+\left(\begin{array}{c}16 \\ 1\end{array}\right)(1)^{15}(-x)^{1}+\left(\begin{array}{c}16 \\ 2\end{array}\right)(1)^{14}(-x)^{2}+\cdots \ldots \ldots$
$+\left(\begin{array}{l}16 \\ 16\end{array}\right)(1)^{0}(-x)^{16}$
$=1-\left(\begin{array}{c}16 \\ 1\end{array}\right) x+\left(\begin{array}{c}16 \\ 2\end{array}\right) x^{2}+\cdots \ldots \ldots+\left(\begin{array}{c}16 \\ 16\end{array}\right) x^{16}$
Now,
$\left(1-3 x+7 x^{2}\right)(1-x)^{16}$
$=\left(1-3 x+7 x^{2}\right)\left(1-\left(\begin{array}{c}16 \\ 1\end{array}\right) x+\left(\begin{array}{c}16 \\ 2\end{array}\right) x^{2}+\cdots \ldots \ldots+\left(\begin{array}{c}16 \\ 16\end{array}\right) x^{16}\right)$
Multiplying the second bracket by $1,(-3 x)$ and $7 x^{2}$
$=\left(1-\left(\begin{array}{c}16 \\ 1\end{array}\right) x+\left(\begin{array}{c}16 \\ 2\end{array}\right) x^{2}+\cdots \ldots \cdots+\left(\begin{array}{c}16 \\ 16\end{array}\right) x^{16}\right)$
$+\left(-3 x+3 x\left(\begin{array}{c}16 \\ 1\end{array}\right) x-3 x\left(\begin{array}{c}16 \\ 2\end{array}\right) x^{2}+\cdots \ldots \ldots-3 x\left(\begin{array}{c}16 \\ 16\end{array}\right) x^{16}\right)$
$+\left(7 x^{2}-7 x^{2}\left(\begin{array}{c}16 \\ 1\end{array}\right) x+7 x^{2}\left(\begin{array}{c}16 \\ 2\end{array}\right) x^{2}+\cdots \ldots \ldots+7 x^{2}\left(\begin{array}{c}16 \\ 16\end{array}\right) x^{16}\right)$
In the above equation terms containing x are
$-\left(\begin{array}{c}16 \\ 1\end{array}\right) x$ and $-3 x$
Therefore, the coefficient of x in the above expansion
$=-\left(\begin{array}{c}16 \\ 1\end{array}\right)-3$
$=-16-3$
$=-19$
$\underline{\text { Conclusion: }}$ coefficient of $x=-19$
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