Question

The coefficient of $x^5$ in the expansion ${(x+3)}^6$ is

• A. 18
• B. 6
• C. 12
• D. 10

Answer 1

The correct option is A 18

As we know that, general term in of expansion ${(a+b)}^n$ is given as-

$T_{r+1}={}^n{C}_r{\left(a\right)}^{n-r}{\left(b\right)}^r$

In the given expansion ${(x+3)}^6$

$a=x,b=3,n=6$

Therefore,

$T_{r+1}={}^6{C}_r{\left(x\right)}^{6-r}{\left(3\right)}^r.....\left(1\right)$

For coefficient of $x^5$,

$6-r=5$

$\Rightarrow r=6-5=1$

Substituting the value of $r$ in ${eq}^n\left(1\right)$, we have

$T_{11+1}={}^6{C}_1{\left(x\right)}^{6-1}{\left(3\right)}^1$

$\Rightarrow T_{12}=18x^5$

Thus the coefficient of $x^5$ is $18$.

Hence the correct answer is $\left(A\right)18$.