Chapter 11.7, Problem 4OE

(a)

To determine

To calculate:

The first terms for the following:

  ${\left(a+b\right)}^{10}$

Answer to Problem 4OE

The first terms in ${\left(a+b\right)}^{10}$ is $a^{10}$

Explanation of Solution

Given information:

  ${\left(a+b\right)}^{10}$

Formula used:

The binomial theorem:

  $\begin{array}{l}{\left(a+b\right)}^n=a^n+n{a}^{n-1}b+\frac{n\left(n-1\right)}{2!}{a}^{n-2}{b}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{a}^{n-3}{b}^3+\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}{a}^{n-4}{b}^4+\mathrm{...}+b^n\\ {\left(a+b\right)}^n=C_0{a}^n{b}^0+C_1{a}^{n-1}{b}^1+C_2{a}^{n-2}{b}^2+C_3{a}^{n-3}{b}^3+C_4{a}^{n-4}{b}^4+\mathrm{...}+C_n{a}^0{b}^n\end{array}$

Calculation:

The given expansion is ${\left(a+b\right)}^{10}$ compared with ${\left(a+b\right)}^n$ , get that: $a=a,b=b,n=10$

The binomial theorem for the given expansion is derived as:

  $\begin{array}{l}{\left(a+b\right)}^{10}=a^{10}+10a^{10-1}b+\mathrm{...}+b^{10}\\ {\left(a+b\right)}^n=a^{10}+10a^{9}b+\mathrm{...}+b^{10}\end{array}$

So, the first term in the above expansion is $a^{10}$

(b)

To determine

To calculate:

The first terms for the following:

  ${\left(3m+n\right)}^3$

Answer to Problem 4OE

The first terms in ${\left(3m+n\right)}^3$ is $27m^3$

Explanation of Solution

Given information:

  ${\left(3m+n\right)}^3$

Formula used:

The binomial theorem:

  $\begin{array}{l}{\left(a+b\right)}^n=a^n+n{a}^{n-1}b+\frac{n\left(n-1\right)}{2!}{a}^{n-2}{b}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{a}^{n-3}{b}^3+\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}{a}^{n-4}{b}^4+\mathrm{...}+b^n\\ {\left(a+b\right)}^n=C_0{a}^n{b}^0+C_1{a}^{n-1}{b}^1+C_2{a}^{n-2}{b}^2+C_3{a}^{n-3}{b}^3+C_4{a}^{n-4}{b}^4+\mathrm{...}+C_n{a}^0{b}^n\end{array}$

Calculation:

The given expansion is ${\left(3m+n\right)}^3$ compared with ${\left(a+b\right)}^n$ , get that: $a=3m,b=n,n=3$

The binomial theorem for the given expansion is derived as:

  $\begin{array}{l}{\left(3m+n\right)}^3={\left(3m\right)}^3+3{\left(3m\right)}^{3-1}n+\mathrm{...}+n^3\\ {\left(3m+n\right)}^3=27m^3+3{\left(3m\right)}^{2}n+\mathrm{...}+n^3\\ {\left(3m+n\right)}^3=27m^3+27m^{2}n+\mathrm{...}+n^3\end{array}$

So, the first term in the above expansion is $27m^3$

(c)

To determine

To calculate:

The first terms for the following:

  ${\left(x^3-y^2\right)}^5$

Answer to Problem 4OE

The first terms in ${\left(x^3-y^2\right)}^5$ is $x^{15}$

Explanation of Solution

Given information:

  ${\left(x^3-y^2\right)}^5$

Formula used:

The binomial theorem:

  $\begin{array}{l}{\left(a+b\right)}^n=a^n+n{a}^{n-1}b+\frac{n\left(n-1\right)}{2!}{a}^{n-2}{b}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{a}^{n-3}{b}^3+\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}{a}^{n-4}{b}^4+\mathrm{...}+b^n\\ {\left(a+b\right)}^n=C_0{a}^n{b}^0+C_1{a}^{n-1}{b}^1+C_2{a}^{n-2}{b}^2+C_3{a}^{n-3}{b}^3+C_4{a}^{n-4}{b}^4+\mathrm{...}+C_n{a}^0{b}^n\end{array}$

Calculation:

The given expansion is ${\left(x^3-y^2\right)}^5$ compared with ${\left(a+b\right)}^n$ , get that: $a=x^3,b=-y^2,n=5$

The binomial theorem for the given expansion is derived as:

  $\begin{array}{l}{\left(x^3-y^2\right)}^5={\left(x^3\right)}^5+5{\left(x^3\right)}^{5-1}\left(-y^2\right)+\mathrm{...}+{\left(-y^2\right)}^5\\ {\left(x^3-y^2\right)}^5=x^{15}-5x^{12}{y}^2+\mathrm{...}+y^{10}\end{array}$

So, the first term in the above expansion is $x^{15}$