Chapter 11.8, Problem 16WE

a.

To determine

To calculate:

The first fourth terms of given expression.

Answer to Problem 16WE

The first fourth terms of ${\left(a+b\right)}^{21}$ are

  $\begin{array}{l}\frac{21!}{0!\left(21-0\right)!}{a}^{21}{b}^0\\ \frac{21!}{1!\left(21-1\right)!}{a}^{20}{b}^1\\ \frac{21!}{2!\left(21-2\right)!}{a}^{19}{b}^2\\ \frac{21!}{3!\left(21-3\right)!}{a}^{18}{b}^3\end{array}$

Explanation of Solution

Given information:

Expression:

  ${\left(a+b\right)}^{21}$

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 ${\left(a+b\right)}^{21}$

Apply the binomial theorem for ${\left(a+b\right)}^{21}$

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

Where,

  $a=a,b=b$

So, the simplification are as follows,

  $\begin{array}{l}{\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i\\ {\left(a+b\right)}^{21}={\displaystyle {\sum }_{i=0}^{21}\left(\begin{array}{c}21\\ i\end{array}\right)}{a}^{\left(21-i\right)}{\left(b\right)}^i\end{array}$

The expand summation are as follows:

  $=\frac{21!}{0!\left(21-0\right)!}{a}^{21}{b}^0+\frac{21!}{1!\left(21-1\right)!}{a}^{20}{b}^1+\mathrm{......}+\frac{21!}{21!\left(21-21\right)!}{a}^0{b}^{21}$

So, the first fourth terms of ${\left(a+b\right)}^{21}$ are as follows,

  $\begin{array}{l}\frac{21!}{0!\left(21-0\right)!}{a}^{21}{b}^0\\ \frac{21!}{1!\left(21-1\right)!}{a}^{20}{b}^1\\ \frac{21!}{2!\left(21-2\right)!}{a}^{19}{b}^2\\ \frac{21!}{3!\left(21-3\right)!}{a}^{18}{b}^3\end{array}$

b.

To determine

To calculate:

The first fourth terms of given expression.

Answer to Problem 16WE

The first fourth terms of ${\left(c-d^2\right)}^{21}$ are,

  $\begin{array}{l}\frac{21!}{0!\left(21-0\right)!}{c}^{21}{\left(-d^2\right)}^0\\ \frac{21!}{1!\left(21-1\right)!}{c}^{20}{\left(-d^2\right)}^1\\ \frac{21!}{2!\left(21-2\right)!}{c}^{19}{\left(-d^2\right)}^2\\ \frac{21!}{3!\left(21-3\right)!}{c}^{18}{\left(-d^2\right)}^3\end{array}$

Explanation of Solution

Given information:

Expression:

  ${\left(c-d^2\right)}^{21}$

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 ${\left(c-d^2\right)}^{21}$

Apply the binomial theorem for ${\left(c-d^2\right)}^{21}$

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

Where,

  $a=c,b=-d^2$

So, the simplification are as follows,

  $\begin{array}{l}{\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i\\ {\left(c-d^2\right)}^{21}={\displaystyle {\sum }_{i=0}^{21}\left(\begin{array}{c}21\\ i\end{array}\right)}{c}^{\left(21-i\right)}{\left(-d^2\right)}^i\end{array}$

The expand summation are as follows:

  $=\frac{21!}{0!\left(21-0\right)!}{c}^{21}{\left(-d^2\right)}^0+\frac{21!}{1!\left(21-1\right)!}{c}^{20}{\left(-d^2\right)}^1+\mathrm{......}+\frac{21!}{21!\left(21-21\right)!}{c}^0{\left(-d^2\right)}^{21}$

So, the first fourth terms of ${\left(c-d^2\right)}^{21}$ are as follows,

  $\begin{array}{l}\frac{21!}{0!\left(21-0\right)!}{c}^{21}{\left(-d^2\right)}^0\\ \frac{21!}{1!\left(21-1\right)!}{c}^{20}{\left(-d^2\right)}^1\\ \frac{21!}{2!\left(21-2\right)!}{c}^{19}{\left(-d^2\right)}^2\\ \frac{21!}{3!\left(21-3\right)!}{c}^{18}{\left(-d^2\right)}^3\end{array}$