Find the coefficient of x in the expansion of

Solution:

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 $x=-19$