Chapter 11.8, Problem 20WE

To determine

To calculate:

Simplify the specified term

Answer to Problem 20WE

The simplification of eighteenth term is $43758s^8{t}^{10}$ .

Explanation of Solution

Given information:

Expression:

  ${\left(s+t\right)}^{18}$

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 ${\left(s+t\right)}^{18}$

Apply the binomial theorem for ${\left(s+t\right)}^{18}$

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

Where,

  $a=s,b=t$

So, the simplification are as follows,

  $\begin{array}{l}{\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i\\ ={\displaystyle {\sum }_{i=0}^{18}\left(\begin{array}{c}18\\ i\end{array}\right)}{s}^{\left(18-i\right)}{{}_{\left(t\right)}}^i\end{array}$

The expand summation are as follows:

  $=\frac{18!}{0!\left(18-0\right)!}{s}^{18}{t}^0+\frac{18!}{1!\left(18-1\right)!}{s}^{18}{t}^1+\mathrm{......}+\frac{18!}{18!\left(18-18\right)!}{s}^0{t}^{18}$

The eighteenth term is,

  $\begin{array}{l}\frac{18!}{10!\left(18-10\right)!}{s}^8{t}^{10}\\ \frac{18!}{10!\left(18-10\right)!}=\frac{1764322560}{8!}\\ =\frac{1764322560}{8!}{s}^8{t}^{10}\end{array}$

Multiply fractions:

  $\begin{array}{l}a.\frac{b}{c}=\frac{a\cdot b}{c}\\ =\frac{1764322560s^8{t}^{10}}{8!}\\ 8!=40320\\ =\frac{1764322560s^8{t}^{10}}{40320}\end{array}$

Divide the numbers:

  $\begin{array}{l}\frac{1764322560s^8{t}^{10}}{40320}\\ =43758\\ =43758s^8{t}^{10}\end{array}$

The answer of eighteenth terms is $43758s^8{t}^{10}$ .