Chapter 11.8, Problem 9OE

a.

To determine

To calculate:

The coefficient of third term of given expression.

Answer to Problem 9OE

The coefficient of the third term of expression of ${\left(a+b\right)}^{10}$ is $45$ .

Explanation of Solution

Given information:

Expression:

  ${\left(a+b\right)}^{10}$

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

  $n=$ total number of terms

Calculation:

Consider the expression ${\left(a+b\right)}^{10}$

From binomial theorem the expression is ${\displaystyle {\sum }_{i=0}^{10}\left(\begin{array}{c}10\\ i\end{array}\right)}{a}^{\left(10-i\right)}{b}^i$

Where,

  $\begin{array}{l}a=a\\ b=b\\ n=10\end{array}$

The expand summation of binomial theorem is,

  $\begin{array}{l}{\left(a+b\right)}^n=C{}_0{a}^n+C{}_1{a}^{n-1}b+C{}_2{a}^{n-2}{b}^2+\mathrm{.....}+C{}_{n-1}a{b}^{n-1}+C{}_n{b}^n\\ {\left(a+b\right)}^{10}=\frac{10!}{0!\left(10-0\right)!}{a}^{10}{b}^0+\mathrm{.......}+\frac{10!}{10!\left(10-10\right)!}{a}^0{b}^{10}\end{array}$

So, from the above theorem the third expression is,

  $\frac{10!}{2!\left(10-2\right)!}{a}^8{b}^2=\frac{90a^8{b}^2}{2}=45a^8{b}^2$

So, the coefficient of the third term of expression of ${\left(a+b\right)}^{10}$ is $45$ .

b.

To determine

To calculate:

The coefficient of third term of given expression.

Answer to Problem 9OE

The coefficient of the third term of expression of ${\left(c-d\right)}^8$ is $28$ .

Explanation of Solution

Given information:

Expression:

  ${\left(c-d\right)}^8$

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

  $n=$ total number of terms

Calculation:

Consider the expression ${\left(c-d\right)}^8$

From binomial theorem the expression is ${\displaystyle {\sum }_{i=0}^8\left(\begin{array}{c}8\\ i\end{array}\right)}{c}^{\left(8-i\right)}{\left(-d\right)}^i$

Where,

  $\begin{array}{l}a=c\\ b=-d\\ n=8\end{array}$

The expand summation of binomial theorem is,

  $\begin{array}{l}{\left(a+b\right)}^n=C{}_0{a}^n+C{}_1{a}^{n-1}b+C{}_2{a}^{n-2}{b}^2+\mathrm{.....}+C{}_{n-1}a{b}^{n-1}+C{}_n{b}^n\\ {\left(c-d\right)}^8=\frac{8!}{0!\left(8-0\right)!}{c}^8{\left(-d\right)}^0+\mathrm{.......}+\frac{8!}{8!\left(8-8\right)!}{c}^0{\left(-d\right)}^8\end{array}$

So, from the above theorem the third expression is,

  $\frac{8!}{2!\left(8-2\right)!}{c}^6{\left(-d\right)}^2=\frac{56c^6{d}^2}{2}=28c^6{d}^2$

So, the coefficient of the third term of expression of ${\left(c-d\right)}^8$ is $28$ .

c.

To determine

To calculate:

The coefficient of third term of given expression.

Answer to Problem 9OE

The coefficient of the third term of expression of ${\left(x+2y\right)}^5$ is $40$ .

Explanation of Solution

Given information:

Expression:

  ${\left(x+2y\right)}^5$

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

  $n=$ total number of terms

Calculation:

Consider the expression ${\left(x+2y\right)}^5$

From binomial theorem the expression is ${\displaystyle {\sum }_{i=0}^5\left(\begin{array}{c}5\\ i\end{array}\right)}{x}^{\left(5-i\right)}{\left(2y\right)}^i$

Where,

  $\begin{array}{l}a=x\\ b=2y\\ n=5\end{array}$

The expand summation of binomial theorem is,

  $\begin{array}{l}{\left(a+b\right)}^n=C{}_0{a}^n+C{}_1{a}^{n-1}b+C{}_2{a}^{n-2}{b}^2+\mathrm{.....}+C{}_{n-1}a{b}^{n-1}+C{}_n{b}^n\\ {\left(x+2y\right)}^5=\frac{5!}{0!\left(5-0\right)!}{x}^5{\left(2y\right)}^0+\mathrm{.......}+\frac{5!}{0!\left(5-0\right)!}{x}^0{\left(2y\right)}^5\end{array}$

So, from the above theorem the third expression is,

  $\frac{5!}{2!\left(5-2\right)!}{x}^3{\left(2y\right)}^2=\frac{80x^3{y}^2}{2!}=40x^3{y}^2$

So, the coefficient of the third term of expression of ${\left(x+2y\right)}^5$ is $40$ .