Chapter 10.5, Problem 65E

To determine

To calculate:

The coefficient of term $x^n$ for $n=4$ from the following binomial expansion:

  ${\left(\frac{1}{4}x-4\right)}^{11}$

Answer to Problem 65E

The coefficient of term $x^n$ for $n=4$ is $-21120$ .

Explanation of Solution

Given information:

  ${\left(\frac{1}{4}x-4\right)}^{11}$

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 ${\left(\frac{1}{4}x-4\right)}^{11}$ compared with ${\left(a+b\right)}^n$ , get that: $a=\frac{1}{4},b=-4,n=4$

The binomial theorem for the given expansion is derived as:

  $\begin{array}{l}{\left(\frac{1}{4}x-4\right)}^{11}=\frac{11!}{0!\left(11-0\right)!}{\left(\frac{x}{4}\right)}^{11}{\left(-4\right)}^0+\frac{11!}{1!\left(11-1\right)!}{\left(\frac{x}{4}\right)}^{10}{\left(-4\right)}^1+\frac{11!}{2!\left(11-2\right)}{\left(\frac{x}{4}\right)}^9{\left(-4\right)}^2+\frac{11!}{3!\left(11-3\right)}{\left(\frac{x}{4}\right)}^8{\left(-4\right)}^3+\frac{11!}{4!\left(11-4\right)}{\left(\frac{x}{4}\right)}^7{\left(-4\right)}^4+\frac{11!}{5!\left(11-5\right)}{\left(\frac{x}{4}\right)}^6{\left(-4\right)}^5+\frac{11!}{6!\left(11-6\right)}{\left(\frac{x}{4}\right)}^5{\left(-4\right)}^6+\frac{11!}{7!\left(11-7\right)}{\left(\frac{x}{4}\right)}^4{\left(-4\right)}^7+\mathrm{...}+\frac{11!}{n!\left(11-n\right)}{\left(\frac{x}{4}\right)}^n{\left(-4\right)}^n\\ {\left(\frac{1}{4}x-4\right)}^{11}=\frac{x^{11}}{4194304}-\frac{x^{10}}{262144}+\frac{55x^9}{16384}-\frac{165x^8}{1024}+\frac{165x^7}{32}-\frac{231x^6}{2}+1848x^5-21120x^4+\mathrm{...}-4194304\end{array}$

Thus, the coefficient of term $x^n$ for $n=4$ is $-21120$ .