Chapter 11.8, Problem 23WE

To determine

To calculate:

Simplify the specified term

Answer to Problem 23WE

The simplification of middle term is

  $1120c^8{d}^4$ .

Explanation of Solution

Given information:

Expression:

  ${\left(c^2-2d\right)}^8$

Formula used:

Binomial theorem:

  ${\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i$

Where,

  $a=$ first term

  $b=$ second term

Calculation:

Consider expression ${\left(c^2-2d\right)}^8$

Apply the binomial theorem for ${\left(c^2-2d\right)}^8$

  ${\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i$

Where,

  $a=c^2,b=-2d$

So, the simplification are as follows,

  $\begin{array}{l}{\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i\\ ={\displaystyle {\sum }_{i=0}^8\left(\begin{array}{c}8\\ i\end{array}\right)}{\left(c^2\right)}^{\left(8-i\right)}{\left(-2d\right)}^i\end{array}$

The expand summation are as follows:

  $=\frac{8!}{0!\left(8-0\right)!}{\left(c^2\right)}^8{\left(-2d\right)}^0+\frac{8!}{1!\left(8-1\right)!}{\left(c^2\right)}^7{\left(-2d\right)}^1+\mathrm{......}+\frac{8!}{8!\left(8-8\right)!}{\left(c^2\right)}^0{\left(-2d\right)}^8$

The middle term is fifth term that is $\frac{8!}{4!\left(8-4\right)!}{\left(c^2\right)}^4{\left(-2d\right)}^4$ ,

  $\begin{array}{l}\frac{8!}{4!\left(8-4\right)!}{\left(c^2\right)}^4{\left(-2d\right)}^4\\ \frac{8!}{4!\left(8-4\right)!}=\frac{1680}{4!}\\ =\frac{1680}{4!}{\left(c^2\right)}^4{\left(-2d\right)}^4\\ {\left(c^2\right)}^4=c^8\\ =\frac{1680}{4!}{c}^8{\left(-2d\right)}^4\\ {\left(-2d\right)}^4=2^4{d}^4\\ =2^4\cdot \frac{1680}{4!}{c}^8{d}^4\end{array}$

Multiply fractions:

  $\begin{array}{l}a\cdot \frac{b}{c}=\frac{a\cdot b}{c}\\ \frac{1680c^8\cdot 2^4{d}^4}{4!}\\ 1680c^8\cdot 2^4{d}^4=26880c^8{d}^4\\ =\frac{26880c^8{d}^4}{4!}\\ 4!=24\\ =\frac{26880c^8{d}^4}{24}\\ \end{array}$

Divide the numbers:

  $\begin{array}{l}\frac{26880c^8{d}^4}{24}\\ =1120\\ =1120c^8{d}^4\end{array}$

The answer of the middle term is $1120c^8{d}^4$ .