Chapter 5.7, Problem 26PPSE

Solution

To identify: The fourth term of ${\left(x+2\right)}^5$ .

The fourth term is $80x^2$ .

Given information:

The given binomial is ${\left(x+2\right)}^5$ .

Explanation:

Consider the given binomial.

In the expansion of the binomial ${\left(a+b\right)}^n$ sum of the powers in each term is equal to the power of the binomial n .

The power of the given binomial is 5 so use the fifth-row of pascal’s triangle.

  

So, the element of the 5th row of the pascal’s triangle are 1, 5, 10, 10, 5, 1. Which is the coefficient of the consecutive terms.

Binomial theorem states that:

  ${\left(a+b\right)}^n=P_0{a}^n+P_1{a}^{n-1}b+\mathrm{...}+P_n{b}^n$ here $P_0,P_1,P_2,P_3,\mathrm{...},P_n$ are the numbers in the nth row of pascal’s triangle.

From the binomial expansion it is clear that the mth term is $P_{m-1}{a}^{n-\left(m-1\right)}{b}^{m-1}$ .

For fourth term $n=5,m=4,a=x\text{ and }b=2$

Thus, the fourth term is:

  $\begin{array}{c}{P}_3{\left(x\right)}^{5-\left(4-1\right)}{\left(2\right)}^{4-1}=10x^2\left(8\right)\\ =80x^2\end{array}$