Chapter 5.7, Problem 22PPSE

Solution

To identify: The expansion of the binomial ${\left(2a+16\right)}^6$ .

The expansion of the binomial is: $64a^6+3,072a^5+61,440a^4+655,360a^3+3,932,160a^2+12,582,912a+16,777,216$

Given information:

The binomial is ${\left(2a+16\right)}^6$ .

Explanation:

Consider the given binomial.

The power of the given binomial is 6 so use the sixth-row of pascal’s triangle.

  

So, the element of the 6th row of the pascal’s triangle are 1, 6, 15, 20, 15, 6, 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.

Replace the coefficient of the consecutive terms.

  $\begin{array}{c}{\left(2a+16\right)}^6=1\cdot {\left(2a\right)}^6{\left(16\right)}^0+6\cdot {\left(2a\right)}^5{\left(16\right)}^1+15\cdot {\left(2a\right)}^4{\left(16\right)}^2+20\cdot {\left(2a\right)}^3{\left(16\right)}^3+15\cdot {\left(2a\right)}^2{\left(16\right)}^4+6\cdot {\left(2a\right)}^1{\left(16\right)}^5\\ +1\cdot {\left(2a\right)}^0{\left(16\right)}^6\\ =64a^6+3,072a^5+61,440a^4+655,360a^3+3,932,160a^2+12,582,912a+16,777,216\end{array}$