Chapter 10.5, Problem 64E

To determine

To calculate:

The coefficient of term $x^n$ for $n=2$ from the following binomial expansion:

  ${\left(3x-1\right)}^9$

Answer to Problem 64E

The coefficient of term $x^n$ for $n=2$ is $-324$ .

Explanation of Solution

Given information:

  ${\left(3x-1\right)}^9$

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

The binomial theorem for the given expansion is derived as:

  $\begin{array}{l}{\left(3x-1\right)}^9=\frac{9!}{0!\left(9-0\right)!}{\left(3x\right)}^9{\left(-1\right)}^0+\frac{9!}{1!\left(9-1\right)!}{\left(3x\right)}^8{\left(-1\right)}^1+\frac{9!}{2!\left(9-2\right)!}{\left(3x\right)}^7{\left(-1\right)}^2+\frac{9!}{3!\left(9-3\right)!}{\left(3x\right)}^6{\left(-1\right)}^3+\frac{9!}{4!\left(9-4\right)!}{\left(3x\right)}^5{\left(-1\right)}^4+\frac{9!}{5!\left(9-5\right)!}{\left(3x\right)}^4{\left(-1\right)}^5+\frac{9!}{6!\left(9-6\right)!}{\left(3x\right)}^3{\left(-1\right)}^6+\frac{9!}{7!\left(9-7\right)!}{\left(3x\right)}^2{\left(-1\right)}^7+\mathrm{...}+\frac{9!}{n!\left(9-n\right)!}{\left(3x\right)}^n{\left(-1\right)}^n\\ {\left(3x-1\right)}^9=19683x^9-59049x^8+78732x^7-61236x^6+30618x^5-10206x^4+2268x^3-324x^2+27x-1\end{array}$

Thus, the coefficient of term $x^n$ for $n=2$ is $-324$ .