Chapter 11.7, Problem 22WE

To determine

To calculate:

The first four terms for the following expansion:

  ${\left(9.8\right)}^4$

Answer to Problem 22WE

The first four terms of ${\left(9.8\right)}^4$ are $10000,-800,24,-0.32$ .

Explanation of Solution

Given information:

  ${\left(9.8\right)}^4$

Formula used:

The binomial theorem using the Pascal’s triangle:

  $\begin{array}{l}{\left(a+b\right)}^n=a^n+n{a}^{n-1}b+\frac{n\left(n-1\right)}{2!}{a}^{n-2}{b}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{a}^{n-3}{b}^3+\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}{a}^{n-4}{b}^4+\mathrm{...}+b^n\\ {\left(a+b\right)}^n=C_0{a}^n{b}^0+C_1{a}^{n-1}{b}^1+C_2{a}^{n-2}{b}^2+C_3{a}^{n-3}{b}^3+C_4{a}^{n-4}{b}^4+\mathrm{...}+C_n{a}^0{b}^n\end{array}$

Calculation:

First need to write the given expansion as simplified form of ${\left(a+b\right)}^n$ :

  ${\left(9.8\right)}^4={\left(10-0.2\right)}^4$

The given expansion is ${\left(10-0.2\right)}^4$ compared with ${\left(a+b\right)}^n$ , get that: $a=10,b=-0.2,n=4$

The binomial theorem for the given expansion is derived as:

  $\begin{array}{l}{\left(10-0.2\right)}^4={\left(10\right)}^4+4{\left(10\right)}^{4-1}\left(-0.2\right)+\frac{4\left(4-1\right)}{2!}{\left(10\right)}^{4-2}{\left(-0.2\right)}^2+\frac{4\left(4-1\right)\left(4-2\right)}{3!}{\left(10\right)}^{4-3}{\left(-0.2\right)}^3+{\left(-0.2\right)}^4\\ {\left(10-0.2\right)}^4=10000-800+24-0.32+1.6\times {10}^{-3}\end{array}$

Therefore, first four terms are $10000,-800,24,-0.32$