Question

Find the coefficient of ${x^5}$ in ${(x + 3)}^8$.

Answer 1

Hint: The knowledge of binomial theorem and expansion is required to solve this problem. The binomial expansion is given by-
${\left( {{\text{a}} + bx} \right)^{\text{n}}} = {}_{}^{\text{n}}{\text{C}}_0^{}{{\text{a}}^{\text{n}}} + {}_{}^{\text{n}}{\text{C}}_1^{}{{\text{a}}^{{\text{n}} - 1}}{{\text{b}}^1}{\text{x}} + ... + {}_{}^{\text{n}}{\text{C}}_{\text{r}}^{}{{\text{a}}^{{\text{n}} - {\text{r}}}}{{\text{b}}^{\text{r}}} + ... + {}_{}^{\text{n}}{\text{C}}_{\text{n}}^{}{{\text{b}}^{\text{n}}}$.

Complete step-by-step solution -
We have to find the coefficient of ${x^5}$ in ${(x + 3)}^8$.
The formula for the ${(r)}^{th}$ general term in a binomial expansion is given as-
${{\text{T}}_{{\text{r}} + 1}} = {}_{}^{\text{n}}{\text{C}}_{\text{r}}^{}{{\text{a}}^{{\text{n}} - {\text{r}}}}{{\text{b}}^{\text{r}}}{{\text{x}}^{\text{r}}}$
Coefficient of $x^5$ = ${}_{}^8{\text{C}}_5^{}{3^{8 - 5}}{1^5}$
Coefficient of $x^5$ = ${}_{}^8{\text{C}}_5^{}{3^3}$
$$=\dfrac{8!}{3!5!}\times 3^3=\dfrac{6\times7\times 8}2\times 3^2\\$$
=1512
Hence, the coefficient of $x^5$ in ${(x + 3)}^8$ is 1512. This is the required answer.

Note: Some students may get ${}_{}^8{\text{C}}_3^{}$ in their formula instead of ${}_{}^8{\text{C}}_5^{}$. But they should not get confused with it. This is a property of combination that-
${}_{}^{\text{n}}{\text{C}}_{\text{r}}^{} = {}_{}^{\text{n}}{\text{C}}_{{\text{n - r}}}^{}$
$$\dfrac{\mathrm n!}{\left(\mathrm n-\mathrm r\right)!\mathrm r!}=\dfrac{\mathrm n!}{\left(\mathrm n-\mathrm n+\mathrm r\right)!\left(\mathrm n-\mathrm r\right)!}$$