# Find the coefficient of x5 in the expansion of

Question:

Find the coefficient of $x^{5}$ in the expansion of $(1+x)^{3}(1-x)^{6}$.

Solution:

To Find : coefficient of $x^{5}$

For $(1+x)^{3}$

$a=1, b=x$ and $n=3$

We have a formula,

$(1+x)^{3}=\sum_{r=0}^{3}\left(\begin{array}{l}3 \\ r\end{array}\right)(1)^{3-r} x^{r}$

$=\left(\begin{array}{l}3 \\ 0\end{array}\right)(1)^{3} x^{0}+\left(\begin{array}{l}3 \\ 1\end{array}\right)(1)^{2} x^{1}+\left(\begin{array}{l}3 \\ 2\end{array}\right)(1)^{1} x^{2}+\left(\begin{array}{l}3 \\ 3\end{array}\right)(1)^{0} x^{3}$

$=1+3 x+3 x^{2}+x^{3}$

For $(1-x)^{6}$

$a=1, b=-x$ and $n=6$

We have formula,

$(1-x)^{6}=\sum_{r=0}^{6}\left(\begin{array}{l}6 \\ r\end{array}\right)(1)^{6-r}(-x)^{r}$

$=\left(\begin{array}{l}6 \\ 0\end{array}\right)(1)^{6}(-x)^{0}+\left(\begin{array}{l}6 \\ 1\end{array}\right)(1)^{5}(-x)^{1}+\left(\begin{array}{l}6 \\ 2\end{array}\right)(1)^{4}(-x)^{2}+\left(\begin{array}{l}6 \\ 3\end{array}\right)(1)^{3}(-x)^{3}$

$+\left(\begin{array}{l}6 \\ 4\end{array}\right)(1)^{2}(-x)^{4}+\left(\begin{array}{l}6 \\ 5\end{array}\right)(1)^{1}(-x)^{5}+\left(\begin{array}{l}6 \\ 6\end{array}\right)(1)^{0}(-x)^{6}$

We have a formula,

$\left(\begin{array}{l}n \\ r\end{array}\right)=\frac{n !}{(n-r) ! \times r !}$

By using this formula, we get, ×

$(1-x)^{6}=1-6 x+15 x^{2}-20 x^{3}+15 x^{4}-6 x^{5}+x^{6}$

$\therefore(1+x)^{3}(1-x)^{6}$

$=\left(1+3 x+3 x^{2}+x^{3}\right)\left(1-6 x+15 x^{2}-20 x^{3}+15 x^{4}-6 x^{5}+x^{6}\right)$

Coefficients of $x^{5}$ are

$x^{0} \cdot x^{5}=1 \times(-6)=-6$

$x^{1} \cdot x^{4}=3 \times 15=45$

$x^{2} \cdot x^{3}=3 \times(-20)=-60$

$x^{3} \cdot x^{2}=1 \times 15=15$

Therefore, Coefficients of $x^{5}=-6+45-60+15=-6$

Conclusion : Coefficients of $x^{5}=-6$