Show that the coefficient of x4 in the expansion of

To show: that the coefficient of $x^{4}$ in the expansion of $\left(1+2 x+x^{2}\right)^{5}$ is 212 .

Formula Used:

We have

$\left(1+2 x+x^{2}\right)^{5}=\left(1+x+x+x^{2}\right)^{5}$

$=(1+x+x(1+x))^{5}$

$=(1+x)^{5}(1+x)^{5}$

$=(1+x)^{10}$

General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by,

$T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where $s$

${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{r !(n-r) !}$

Now, finding the general term,

$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{Cr} \times \mathrm{X}^{10-\mathrm{r}} \times(1)^{\mathrm{r}}$

$10-r=4$

$r=6$

Thus, the coefficient of $x^{4}$ in the expansion of $\left(1+2 x+x^{2}\right)^{5}$ is given by,

${ }^{10} \mathrm{C}_{4}=\frac{10 !}{4 ! 6 !}$

${ }^{10} \mathrm{C}_{4}=\frac{10 \times 9 \times 8 \times 7 \times 6 !}{24 \times 6 !}$

${ }^{10} \mathrm{C}_{4}=210$

Thus, the coefficient of $x^{4}$ in the expansion of $\left(1+2 x+x^{2}\right)^{5}$ is 210