Question

Find the coefficient of $x^2$in the expansion of${\left(2x+1\right)}^7$.

Short answer

The coefficient of $x^2$is$84.$

Step-by-step solution

Given information:

Consider the expression ${\left(2x+1\right)}^7$.

To find the coefficient of $x^2$

Formulas :

Formula 1: To find a particular term in an expression:

${\left(x+a\right)}^n=\left(\begin{array}{c}n\\ n-j\end{array}\right)a^{(n-j)}{x}^j$

Formula 2: Binomial Coefficient:

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

Step 3:

On comparing ${\left(2x+1\right)}^7$with ${\left(x+a\right)}^n$, we get,

$x=2x,a=1,n=7$

To find coefficient of$x^2$,$\Rightarrow j=2$.

Substituting the values in the formula 1:

${\left(2x+1\right)}^7=\left(\begin{array}{c}7\\ 7-2\end{array}\right){\left(1\right)}^{(7-2)}{\left(2x\right)}^2$

Simplification:

${\left(2x+1\right)}^7=\left(\begin{array}{c}7\\ 5\end{array}\right)1^5\left(2^2{x}^2\right)$

$=21\times 4x^2$

role="math" localid="1647343658235" $\left[Useformula2:\left(\begin{array}{c}7\\ 5\end{array}\right)=\frac{7!}{5!\times \left(7-5\right)!}=\frac{7!}{5!\times 2!}=\frac{7\times 6\times 5!}{5!\times (2\times 1)}=\frac{7\times 6}{2}=21\right]$

$=84x^2$