Chapter 9.2, Problem 7E

Solution

To identify: The expansion of the binomial ${\left(p+q\right)}^8$ by using Pascal’s triangle.

The expanded form of the binomial is

  ${\left(p+q\right)}^8=p^8+8p^{7}q+28p^6{q}^2+56p^5{q}^3+70p^4{q}^4+56p^3{q}^5+28p^2{q}^6+8p{q}^7+q^8$

Given information:

The binomial is ${\left(p+q\right)}^8$ .

Explanation:

Consider the given binomial

The power of the given binomial is $8$ so use the eighth -row of pascal's triangle.

  

So, the element of the $8$ th row of the pascal's triangle are $\left(1,8,28,56,70,56,28,8,1\right)$ Which is the coefficient of the consecutive terms.

Binomial theorem states that:

  ${\left(a+b\right)}^n=P_0{\text{a}}^1+P_1{a}^{n-1}b+\mathrm{....}+P_n{\text{b}}^n\text{here}{P}_0,P_1,P_2,P_3,P_n$ are the numbers in the nth row

of pascal’s triangle.

Replace the coefficient of the consecutive terms.

  $\begin{array}{c}{\left(p+q\right)}^8=1{\left(p\right)}^8+8{\left(p\right)}^7\left(q\right)+28{\left(p\right)}^6{\left(q\right)}^2+56{\left(p\right)}^5{\left(q\right)}^3+70{\left(p\right)}^4{\left(q\right)}^4\\ +56{\left(p\right)}^3{\left(q\right)}^5+28{\left(p\right)}^2{\left(q\right)}^6+8\left(p\right){\left(q\right)}^7+{\left(q\right)}^8\\ {\left(p+q\right)}^8=p^8+8p^{7}q+28p^6{q}^2+56p^5{q}^3+70p^4{q}^4+56p^3{q}^5+28p^2{q}^6+8p{q}^7+q^8\end{array}$