Chapter 12, Problem 38SGA

To determine

To find: the 3^rd term of the binomial.

Answer to Problem 38SGA

  $114688m^6$

Explanation of Solution

Given information:

  ${\left(4m+1\right)}^8$

Calculation:

For positive integer $n$ , then the following will hold true.

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\frac{n\left(n-1\right)x^{n-2}{y}^2}{\left(1\right)\left(2\right)}+\frac{n\left(n-1\right)\left(n-2\right)x^{n-3}{y}^3}{\left(1\right)\left(2\right)}+\mathrm{...}+y^n$

The coefficients ${}_n{\text{C}}_k$ can be written as ${}_n{\text{C}}_k=\frac{n!}{k!\left(n-k\right)!}$ .

Let the 3^rd term of the binomial be $z$ . So put $n=8,k=2$ and solve as $z$ follows:

  $\begin{array}{c}z=\frac{8!}{2!\left(8-2\right)!}\cdot {\left(4m\right)}^{8-2}\cdot {\left(1\right)}^2\\ =28\left(4096m^6\right)\left(1\right)\\ =114688m^6\end{array}$

Thus the 3^rd term of the binomial is $114688m^6$ .