Chapter 11.7, Problem 5OE

(a)

To determine

To calculate:

The last term for the following:

  ${\left(a-b\right)}^7$

Answer to Problem 5OE

The last term in ${\left(a-b\right)}^7$ is $-b^7$

Explanation of Solution

Given information:

  ${\left(a-b\right)}^7$

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 ${\left(a-b\right)}^7$ compared with ${\left(a+b\right)}^n$ , get that: $a=a,b=-b,n=7$

The binomial theorem for the given expansion is derived as:

  $\begin{array}{l}{\left(a-b\right)}^7=a^7+7a^{7-1}\left(-b\right)+\mathrm{...}+{\left(-b\right)}^7\\ {\left(a-b\right)}^7=a^7-7a^{6}b+\mathrm{...}-b^7\end{array}$

So, the last term in the above expansion is $-b^7$

(b)

To determine

To calculate:

The last term for the following:

  ${\left(p^2-2\right)}^4$

Answer to Problem 5OE

The last term in ${\left(p^2-2\right)}^4$ is $27m^3$

Explanation of Solution

Given information:

  ${\left(p^2-2\right)}^4$

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 ${\left(p^2-2\right)}^4$ compared with ${\left(a+b\right)}^n$ , get that: $a=p^2,b=-2,n=4$

The binomial theorem for the given expansion is derived as:

  $\begin{array}{l}{\left(p^2-2\right)}^4={\left(p^2\right)}^4+4{\left(p^2\right)}^{4-1}\left(-2\right)+\mathrm{...}+{\left(-2\right)}^4\\ {\left(p^2-2\right)}^4=p^8-8p^6+\mathrm{...}+16\end{array}$

So, the last term in the above expansion is $16$

(c)

To determine

To calculate:

The last term for the following:

  ${\left(m+n^2\right)}^8$

Answer to Problem 5OE

The last term in ${\left(m+n^2\right)}^8$ is $n^{16}$

Explanation of Solution

Given information:

  ${\left(m+n^2\right)}^8$

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 ${\left(m+n^2\right)}^8$ compared with ${\left(a+b\right)}^n$ , get that: $a=m,b=n^2,n=8$

The binomial theorem for the given expansion is derived as:

  $\begin{array}{l}{\left(m+n^2\right)}^8={\left(m\right)}^8+8{\left(m\right)}^{8-1}\left(n^2\right)+\mathrm{...}+{\left(n^2\right)}^8\\ {\left(m+n^2\right)}^8=m^8+8m^7{n}^2+\mathrm{...}+n^{16}\end{array}$

So, the first term in the above expansion is $n^{16}$