# Find the coefficient of x4 in the expansion of

Question:

Find the coefficient of $x^{4}$ in the expansion of $(1+x)^{n}(1-x)^{n}$. Deduce that $C_{2}=C_{0} C_{4}-C_{1} C_{3}+C_{2} C_{2}-C_{3} C_{1}+C_{4} C_{0}$, where $C_{r}$ stands for ${ }^{n} C_{r}$

Solution:

To Find : Coefficients of $x^{4}$

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

a=1, b=x

We have a formula,

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

$=\left(\begin{array}{l}\mathrm{n} \\ 0\end{array}\right)(1)^{\mathrm{n}} \mathrm{x}^{0}+\left(\begin{array}{l}\mathrm{n} \\ 1\end{array}\right)(1)^{\mathrm{n}-1} \mathrm{x}^{1}+\left(\begin{array}{l}\mathrm{n} \\ 2\end{array}\right)(1)^{\mathrm{n}-2} \mathrm{x}^{2}+\cdots+\left(\begin{array}{l}\mathrm{n} \\ \mathrm{n}\end{array}\right)(1)^{\mathrm{n}-\mathrm{n}} \mathrm{x}^{\mathrm{n}}$

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

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

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

We have formula,

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

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

$+\left(\begin{array}{l}\mathrm{n} \\ \mathrm{n}\end{array}\right)(1)^{\mathrm{n}-\mathrm{n}}(-\mathrm{x})^{\mathrm{n}}$

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

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

$=\left\{\left(\begin{array}{l}n \\ 0\end{array}\right) x^{0}+\left(\begin{array}{l}n \\ 1\end{array}\right) x+\left(\begin{array}{l}n \\ 2\end{array}\right) x^{2}+\cdots+\left(\begin{array}{l}n \\ n\end{array}\right) x^{n}\right\}\left\{\left(\begin{array}{l}n \\ 0\end{array}\right)(-x)^{0}-\left(\begin{array}{l}n \\ 1\end{array}\right)(x)^{1}+\left(\begin{array}{l}n \\ 2\end{array}\right)(x)^{2}\right.$

$\left.+\cdots+\left(\begin{array}{l}n \\ n\end{array}\right)(-x)^{n}\right\}$

Coefficients of $x^{4}$ are

$x^{0} \cdot x^{4}=\left(\begin{array}{l}n \\ 0\end{array}\right) \times\left(\begin{array}{l}n \\ 4\end{array}\right)=C_{0} C_{4}$

$x^{1} \cdot x^{3}=\left(\begin{array}{l}n \\ 1\end{array}\right) \times(-1)\left(\begin{array}{l}n \\ 3\end{array}\right)=-\left(\begin{array}{l}n \\ 1\end{array}\right)\left(\begin{array}{l}n \\ 3\end{array}\right)=-C_{1} C_{3}$

$x^{2} \cdot x^{2}=\left(\begin{array}{l}\mathrm{n} \\ 2\end{array}\right) \times\left(\begin{array}{l}\mathrm{n} \\ 2\end{array}\right)=C_{2} C_{2}$

$x^{3} \cdot x^{1}=\left(\begin{array}{l}n \\ 3\end{array}\right) \times(-1)\left(\begin{array}{l}n \\ 1\end{array}\right)=-\left(\begin{array}{l}n \\ 3\end{array}\right)\left(\begin{array}{l}n \\ 1\end{array}\right)=-C_{3} C_{1}$

$x^{4} \cdot x^{0}=\left(\begin{array}{l}n \\ 4\end{array}\right) \times\left(\begin{array}{l}n \\ 0\end{array}\right)=C_{4} C_{0}$

Therefore, Coefficient of $x^{4}$

$=C_{4} C_{0}-C_{1} C_{3}+C_{2} C_{2}-C_{3} C_{1}+C_{4} C_{0}$

Let us assume, n=4, it becomes

${ }^{4} \mathrm{C}_{4}{ }^{4} \mathrm{C}_{0}-{ }^{4} \mathrm{C}_{1}{ }^{4} \mathrm{C}_{3}+{ }^{4} \mathrm{C}_{2}{ }^{4} \mathrm{C}_{2}-{ }^{4} \mathrm{C}_{3}{ }^{4} \mathrm{C}_{1}+{ }^{4} \mathrm{C}_{4}{ }^{4} \mathrm{C}_{0}$

We know that

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

By using above formula, we get,

${ }^{4} \mathrm{C}_{4}{ }^{4} \mathrm{C}_{0}-{ }^{4} \mathrm{C}_{1}{ }^{4} \mathrm{C}_{3}+{ }^{4} \mathrm{C}_{2}{ }^{4} \mathrm{C}_{2}-{ }^{4} \mathrm{C}_{3}{ }^{4} \mathrm{C}_{1}+{ }^{4} \mathrm{C}_{4}{ }^{4} \mathrm{C}_{0}$

$=(1)(1)-(4)(4)+(6)(6)-(4)(4)+(1)(1)$

$=1-16+36-16+1$

$=6$

$={ }^{4} \mathrm{C} 2$

Therefore, in general,

$\mathrm{C}_{4} \mathrm{C}_{0}-\mathrm{C}_{1} \mathrm{C}_{3}+\mathrm{C}_{2} \mathrm{C}_{2}-\mathrm{C}_{3} \mathrm{C}_{1}+\mathrm{C}_{4} \mathrm{C}_{0}=\mathrm{C}_{2}$

Therefore, Coefficient of $x^{4}=C_{2}$

Conclusion :

- Coefficient of $x^{4}=C_{2}$

$\cdot C_{4} C_{0}-C_{1} C_{3}+C_{2} C_{2}-C_{3} C_{1}+C_{4} C_{0}=C_{2}$