Chapter 11.8, Problem 10OE

a.

To determine

To calculate:

The power of the fifth terms $a$ of given expression.

Answer to Problem 10OE

The power of $a$ is $10$ .

Explanation of Solution

Given information:

Expression:

  ${\left(a+b\right)}^{20}$

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,

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

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 fifth term is as follows,

  $\begin{array}{l}\frac{20!}{4!\left(20-4\right)!}{a}^{16}{b}^4\\ \frac{20!}{4!\left(20-4\right)!}=\frac{116280}{4!}\\ =\frac{116280}{4!}{a}^{16}{b}^4\end{array}$

Multiply fractions:

  $\begin{array}{l}a.\frac{b}{c}=\frac{a\cdot b}{c}\\ =\frac{116280a^{16}{b}^4}{4!}\\ 4!=24\\ =\frac{116280a^{16}{b}^4}{24}\end{array}$

Divide the numbers:

  $\begin{array}{l}\frac{116280}{24}\\ =4845a^{16}{b}^4\end{array}$

The fifth term is $4845a^{16}{b}^4$ and the power of $a$ is $10$ .

b.

To determine

To calculate:

The power of the eleventh terms $b$ of given expression.

Answer to Problem 10OE

The power of $b$ is $10$ .

Explanation of Solution

Given information:

Expression:

  ${\left(a+b\right)}^{20}$

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,

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

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 eleventh term is as follows,

  $\begin{array}{l}\frac{20!}{10!\left(20-10\right)!}{a}^{10}{b}^{10}\\ \frac{20!}{10!\left(20-10\right)!}=\frac{670442572800}{10!}\\ =\frac{670442572800}{10!}{a}^{10}{b}^{10}\end{array}$

Multiply fractions:

  $\begin{array}{l}a.\frac{b}{c}=\frac{a\cdot b}{c}\\ =\frac{670442572800}{10!}{a}^{10}{b}^{10}\\ =10!\\ =3628800\\ =\frac{670442572800a^{10}{b}^{10}}{3628800}\end{array}$

Divide the numbers:

  $\frac{670442572800}{3628800}=184756=184756a^{10}{b}^{10}$

The eleventh term is $184756a^{10}{b}^{10}$ and the power of $b$ is $10$ .