Chapter 11.8, Problem 15WE

a.

To determine

To calculate:

The first fourth terms of given expression.

Answer to Problem 15WE

The first fourth terms of ${\left(a+b\right)}^{19}$ are

  $\begin{array}{l}\frac{19!}{0!\left(19-0\right)!}{a}^{19}{b}^0\\ \frac{19!}{1!\left(19-1\right)!}{a}^{18}{b}^1\\ \frac{19!}{2!\left(19-2\right)!}{a}^{17}{b}^2\\ \frac{19!}{3!\left(19-3\right)!}{a}^{16}{b}^3\end{array}$

Explanation of Solution

Given information:

Expression:

  ${\left(a+b\right)}^{19}$

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 ${\left(a+b\right)}^{19}$

Apply the binomial theorem for ${\left(a+b\right)}^{19}$

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

Where,

  $a=a,b=b$

So, the simplification are as follows,

  $\begin{array}{l}{\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i\\ {\left(a+b\right)}^{19}={\displaystyle {\sum }_{i=0}^{19}\left(\begin{array}{c}19\\ i\end{array}\right)}{a}^{\left(19-i\right)}{\left(b\right)}^i\end{array}$

The expand summation are as follows:

  $=\frac{19!}{0!\left(19-0\right)!}{a}^{19}{b}^0+\frac{19!}{1!\left(19-1\right)!}{a}^{18}{b}^1+\mathrm{......}+\frac{19!}{19!\left(19-19\right)!}{a}^0{b}^{19}$

So, the first fourth terms of ${\left(a+b\right)}^{19}$ are as follows,

  $\begin{array}{l}\frac{19!}{0!\left(19-0\right)!}{a}^{19}{b}^0\\ \frac{19!}{1!\left(19-1\right)!}{a}^{18}{b}^1\\ \frac{19!}{2!\left(19-2\right)!}{a}^{17}{b}^2\\ \frac{19!}{3!\left(19-3\right)!}{a}^{16}{b}^3\end{array}$

b.

To determine

To calculate:

The first fourth terms of given expression.

Answer to Problem 15WE

The first fourth terms of ${\left(x-2y\right)}^{19}$ are,

  $\begin{array}{l}\frac{19!}{0!\left(19-0\right)!}{x}^{19}{\left(-2y\right)}^0\\ \frac{19!}{1!\left(19-1\right)!}{x}^{18}{\left(-2y\right)}^1\\ \frac{19!}{2!\left(19-2\right)!}{x}^{17}{\left(-2y\right)}^2\\ \frac{19!}{3!\left(19-3\right)!}{x}^{16}{\left(-2y\right)}^3\end{array}$

Explanation of Solution

Given information:

Expression:

  ${\left(x-2y\right)}^{19}$

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 ${\left(x-2y\right)}^{19}$

Apply the binomial theorem for ${\left(x-2y\right)}^{19}$

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

Where,

  $a=x,b=-2y$

So, the simplification are as follows,

  $\begin{array}{l}{\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i\\ {\left(x-2y\right)}^{19}={\displaystyle {\sum }_{i=0}^{19}\left(\begin{array}{c}19\\ i\end{array}\right)}{a}^{\left(19-i\right)}{\left(-2y\right)}^i\end{array}$

The expand summation are as follows:

  $=\frac{19!}{0!\left(19-0\right)!}{x}^{19}{\left(-2y\right)}^0+\frac{19!}{1!\left(19-1\right)!}{x}^{18}{\left(-2y\right)}^1+\mathrm{......}+\frac{19!}{19!\left(19-19\right)!}{x}^0{\left(-2y\right)}^{19}$

So, the first fourth terms of ${\left(x-2y\right)}^{19}$ are as follows,

  $\begin{array}{l}\frac{19!}{0!\left(19-0\right)!}{x}^{19}{\left(-2y\right)}^0\\ \frac{19!}{1!\left(19-1\right)!}{x}^{18}{\left(-2y\right)}^1\\ \frac{19!}{2!\left(19-2\right)!}{x}^{17}{\left(-2y\right)}^2\\ \frac{19!}{3!\left(19-3\right)!}{x}^{16}{\left(-2y\right)}^3\end{array}$