Chapter 11.7, Problem 3OE

(a)

To determine

To calculate:

The number of terms for the following:

  ${\left(m-n\right)}^1$

Answer to Problem 3OE

The two terms are available in ${\left(m-n\right)}^1$ .

Explanation of Solution

Given information:

  ${\left(m-n\right)}^1$

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 ${\left(m-n\right)}^1$ compared with ${\left(a+b\right)}^n$ , get that: $a=m,b=-n,n=1$

The binomial theorem for the given expansion is derived as:

  $\begin{array}{l}{\left(m-n\right)}^1=C{}_0{m}^1{\left(-n\right)}^0+C{}_1{\left(m\right)}^{1-1}{\left(-n\right)}^1\\ {\left(m-n\right)}^1=\frac{1!}{\left(1-0\right)!0!}m\left(1\right)+\frac{1!}{\left(1-1\right)!1!}{m}^0\left(-n\right)\left[\because x^0=1\right]\\ {\left(m-n\right)}^1=\left(1\right)m\left(1\right)+\left(1\right)\left(1\right)\left(-n\right)\\ {\left(m-n\right)}^1=m-n\end{array}$

Hence, two terms are available in the given expansion.

(b)

To determine

To calculate:

The number of terms for the following:

  ${\left(2x+y\right)}^6$

Answer to Problem 3OE

The seven terms are available in ${\left(2x+y\right)}^6$ .

Explanation of Solution

Given information:

  ${\left(2x+y\right)}^6$

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 ${\left(2x+y\right)}^6$ compared with ${\left(a+b\right)}^n$ , get that: $a=2x,b=y,n=6$

The binomial theorem for the given expansion is derived as:

  ${\left(a+b\right)}^n=a^n+n{a}^{n-1}b+\frac{n\left(n-1\right)}{2!}{a}^{n-2}{b}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{a}^{n-3}{b}^3+\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}{a}^{n-4}{b}^4+\mathrm{...}+b^n$ put the given data:

  $\begin{array}{l}{\left(2x+y\right)}^6={\left(2x\right)}^6+6{\left(2x\right)}^{6-1}y+\frac{6\left(6-1\right)}{2!}{\left(2x\right)}^{6-2}{y}^2+\frac{6\left(6-1\right)\left(6-2\right)}{3!}{\left(2x\right)}^{6-3}{y}^3\\ +\frac{6\left(6-1\right)\left(6-2\right)\left(6-3\right)}{4!}{\left(2x\right)}^{6-4}{y}^4+\frac{6\left(6-1\right)\left(6-2\right)\left(6-3\right)\left(6-4\right)}{5!}{\left(2x\right)}^{6-5}{y}^5\\ +\frac{6\left(6-1\right)\left(6-2\right)\left(6-3\right)\left(6-4\right)\left(6-5\right)}{6!}{\left(2x\right)}^{6-6}{b}^6\\ {\left(2x+y\right)}^6=64x^6+192x^{5}y+240x^4{y}^2+160x^3{y}^3+60x^2{y}^4+12x{y}^5+y^6\end{array}$

Hence, seven terms are available in the given expansion.

(c)

To determine

To calculate:

The number of terms for the following:

  ${\left(3p^2-1\right)}^9$

Answer to Problem 3OE

The ten terms are available in ${\left(3p^2-1\right)}^9$ .

Explanation of Solution

Given information:

  ${\left(3p^2-1\right)}^9$

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 ${\left(3p^2-1\right)}^9$ compared with ${\left(a+b\right)}^n$ , get that: $a=3p^2,b=1,n=9$

The binomial theorem for the given expansion is derived as:

  ${\left(a+b\right)}^n=a^n+n{a}^{n-1}b+\frac{n\left(n-1\right)}{2!}{a}^{n-2}{b}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{a}^{n-3}{b}^3+\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}{a}^{n-4}{b}^4+\mathrm{...}+b^n$ put the given data:

  ${\left(a+b\right)}^n=a^n+n{a}^{n-1}b+\frac{n\left(n-1\right)}{2!}{a}^{n-2}{b}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{a}^{n-3}{b}^3+\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}{a}^{n-4}{b}^4+\mathrm{...}+b^n$

  ${\left(3p^2-1\right)}^9={\left(3p^2\right)}^n+9{\left(3p^2\right)}^{9-1}b+\frac{9\left(9-1\right)}{2!}{\left(3p^2\right)}^{9-2}{b}^2+\frac{9\left(9-1\right)\left(9-2\right)}{3!}{\left(3p^2\right)}^{9-3}{b}^3$

  $+\frac{9\left(9-1\right)\left(9-2\right)\left(9-3\right)}{4!}{\left(3p^2\right)}^{9-4}{b}^4$

  $+\frac{9\left(9-1\right)\left(9-2\right)\left(9-3\right)\left(9-4\right)}{5!}{\left(3p^2\right)}^{9-5}{b}^5$

  $+\frac{9\left(9-1\right)\left(9-2\right)\left(9-3\right)\left(9-4\right)\left(9-5\right)}{6!}{\left(3p^2\right)}^{9-6}{b}^6$

  $+\frac{9\left(9-1\right)\left(9-2\right)\left(9-3\right)\left(9-4\right)\left(9-5\right)\left(9-6\right)}{7!}{\left(3p^2\right)}^{9-7}{b}^7$

  $+\frac{9\left(9-1\right)\left(9-2\right)\left(9-3\right)\left(9-4\right)\left(9-5\right)\left(9-6\right)\left(9-7\right)}{8!}{\left(3p^2\right)}^{9-8}{b}^8$

  $+\frac{9\left(9-1\right)\left(9-2\right)\left(9-3\right)\left(9-4\right)\left(9-5\right)\left(9-6\right)\left(9-7\right)\left(9-8\right)}{9!}{\left(3p^2\right)}^{9-9}{b}^9$

  ${\left(3p^2-1\right)}^9=19683p^{18}-59049p^{16}+78732p^{14}-61236p^{12}+30618p^{10}-10206p^8+2268p^6-324p^4+27p^2-1$

Hence, ten terms are available in the given expansion.