Chapter 11.8, Problem 14WE

a.

To determine

To calculate:

The first fourth terms of given expression.

Answer to Problem 14WE

The first fourth terms of ${\left(a+b\right)}^{12}$ are

  $\begin{array}{l}\frac{12!}{0!\left(12-0\right)!}{a}^{12}{b}^0\\ \frac{12!}{1!\left(12-1\right)!}{a}^{11}{b}^1\\ \frac{12!}{2!\left(12-2\right)!}{a}^{10}{b}^2\\ \frac{12!}{3!\left(12-3\right)!}{a}^9{b}^3\end{array}$

Explanation of Solution

Given information:

Expression:

  ${\left(a+b\right)}^{12}$

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 ${\left(a+b\right)}^{12}$

Apply the binomial theorem for ${\left(a+b\right)}^{12}$

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

Where,

  $a=a,b=b$

So, the simplification are as follows,

  $\begin{array}{l}{\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i\\ {\left(a+b\right)}^{12}={\displaystyle {\sum }_{i=0}^{12}\left(\begin{array}{c}12\\ i\end{array}\right)}{a}^{\left(12-i\right)}{\left(b\right)}^i\end{array}$

The expand summation are as follows:

  $=\frac{12!}{0!\left(12-0\right)!}{a}^{12}{b}^0+\frac{12!}{1!\left(12-1\right)!}{a}^{11}{b}^1+\mathrm{......}+\frac{12!}{12!\left(12-12\right)!}{a}^0{b}^{12}$

So, the first fourth terms of ${\left(a+b\right)}^{12}$ are as follows,

  $\begin{array}{l}\frac{12!}{0!\left(12-0\right)!}{a}^{12}{b}^0\\ \frac{12!}{1!\left(12-1\right)!}{a}^{11}{b}^1\\ \frac{12!}{2!\left(12-2\right)!}{a}^{10}{b}^2\\ \frac{12!}{3!\left(12-3\right)!}{a}^9{b}^3\end{array}$

b.

To determine

To calculate:

The first fourth terms of given expression.

Answer to Problem 14WE

The first fourth terms of ${\left(a-b\right)}^{12}$ are,

  $\begin{array}{l}\frac{12!}{0!\left(12-0\right)!}{a}^{12}{\left(-b\right)}^0\\ \frac{12!}{1!\left(12-1\right)!}{a}^{11}{\left(-b\right)}^1\\ \frac{12!}{2!\left(12-2\right)!}{a}^{10}{\left(-b\right)}^2\\ \frac{12!}{3!\left(12-3\right)!}{a}^9{\left(-b\right)}^3\end{array}$

Explanation of Solution

Given information:

Expression:

  ${\left(a-b\right)}^{12}$

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 ${\left(a-b\right)}^{12}$

Apply the binomial theorem for ${\left(a-b\right)}^{12}$

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

Where,

  $a=a,b=-b$

So, the simplification are as follows,

  $\begin{array}{l}{\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i\\ {\left(a-b\right)}^{12}={\displaystyle {\sum }_{i=0}^{12}\left(\begin{array}{c}12\\ i\end{array}\right)}{a}^{\left(12-i\right)}{\left(-b\right)}^i\end{array}$

The expand summation are as follows:

  $=\frac{12!}{0!\left(12-0\right)!}{a}^{12}{\left(-b\right)}^0+\frac{12!}{1!\left(12-1\right)!}{a}^{11}{\left(-b\right)}^1+\mathrm{......}+\frac{12!}{12!\left(12-12\right)!}{a}^0{\left(-b\right)}^{12}$

So, the first fourth terms of ${\left(a-b\right)}^{12}$ are as follows,

  $\begin{array}{l}\frac{12!}{0!\left(12-0\right)!}{a}^{12}{\left(-b\right)}^0\\ \frac{12!}{1!\left(12-1\right)!}{a}^{11}{\left(-b\right)}^1\\ \frac{12!}{2!\left(12-2\right)!}{a}^{10}{\left(-b\right)}^2\\ \frac{12!}{3!\left(12-3\right)!}{a}^9{\left(-b\right)}^3\end{array}$