Chapter 10.5, Problem 54E

To determine

To calculate:

The number of terms for the following:

  ${\left(4p-q\right)}^6$

Answer to Problem 54E

The binomial expansion is:

  $4096p^6-6144p^{5}q+3840p^4{q}^2-1280p^3{q}^3+240p^2{q}^2-24p{q}^5+q^6$ .

Explanation of Solution

Given information:

  ${\left(4p-q\right)}^6$

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 ${\left(4p-q\right)}^6$ compared with ${\left(a+b\right)}^n$ , get that: $a=4p,b=q,n=6$

The binomial theorem for the given expansion is derived as:

Hence, the binomial expansion is:

  $4096p^6-6144p^{5}q+3840p^4{q}^2-1280p^3{q}^3+240p^2{q}^2-24p{q}^5+q^6$ .