Chapter 11.8, Problem 17WE

To determine

To calculate:

Simplify the specified term

Answer to Problem 17WE

The simplification of term that containing $b^5$ is $15504a^{15}{b}^5$ .

Explanation of Solution

Given information:

Expression:

  ${\left(a+b\right)}^{20}$

Term:

  $b^5$

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 ${\left(a+b\right)}^{20}$

Apply the binomial theorem for ${\left(a+b\right)}^{20}$

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

Where,

  $a=a,b=b$

So, the simplification are as follows,

  $\begin{array}{l}{\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i\\ ={\displaystyle {\sum }_{i=0}^{20}\left(\begin{array}{c}20\\ i\end{array}\right)}{x}^{\left(20-i\right)}{b}^i\end{array}$

The expand summation are as follows:

  $=\frac{20!}{0!\left(20-0\right)!}{a}^{20}{b}^0+\frac{20!}{1!\left(20-1\right)!}{a}^{19}{b}^1+\mathrm{......}+\frac{20!}{20!\left(20-20\right)!}{a}^0{b}^{20}$

The term that contains the $b^5$ is the sixth term,

  $\begin{array}{l}\frac{20!}{5!\left(20-5\right)!}{a}^{15}{b}^5\\ =\frac{1860480}{5!}\\ =\frac{1860480}{5!}{a}^{15}{b}^5\end{array}$

Multiply fractions:

  $\begin{array}{l}a.\frac{b}{c}=\frac{a\cdot b}{c}\\ =-\frac{1860480a^{15}{b}^5}{120}\\ 5!=120\\ =\frac{1860480a^{15}{b}^5}{120}\end{array}$

Divide the numbers:

  $\begin{array}{l}\frac{1860480}{120}\\ =15504\\ =15504a^{15}{b}^5\end{array}$

The answer of term that containing the $b^5$ is $15504a^{15}{b}^5$ .