Chapter 10.5, Problem 61E

To determine

To calculate:

The coefficient of term $x^n$ for $n=6$ from the following binomial expansion:

  ${\left(x^2-3\right)}^8$

Answer to Problem 61E

The coefficient of term $x^n$ for $n=6$ is $-13608$ .

Explanation of Solution

Given information:

  ${\left(x^2-3\right)}^8$

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 ${\left(x^2-3\right)}^8$ compared with ${\left(a+b\right)}^n$ , get that: $a=x^2,b=-3,n=8$

The binomial theorem for the given expansion is derived as:

  $\begin{array}{l}{\left(x^2-3\right)}^8={\left(x^2\right)}^8+\left(8\right){\left(x^2\right)}^{8-1}\left(-3\right)+\frac{8\left(8-1\right)}{2!}{\left(x^2\right)}^{8-2}{\left(-3\right)}^2+\frac{8\left(8-1\right)\left(8-2\right)}{3!}{\left(x^2\right)}^{8-3}{\left(-3\right)}^3\\ +\frac{8\left(8-1\right)\left(8-2\right)\left(8-3\right)}{4!}{\left(x^2\right)}^{8-4}{\left(-3\right)}^4+\frac{8\left(8-1\right)\left(8-2\right)\left(8-3\right)\left(8-4\right)}{5!}{\left(x^2\right)}^{8-5}{\left(-3\right)}^5\\ +\frac{8\left(8-1\right)\left(8-2\right)\left(8-3\right)\left(8-4\right)\left(8-5\right)}{6!}{\left(x^2\right)}^{8-6}{\left(-3\right)}^6+\mathrm{...}+{\left(-3\right)}^8\\ {\left(x^2-3\right)}^8=x^{16}-24x^{14}+252x^{12}-1512x^{10}+5670x^8-13608x^6+\mathrm{...}+6561\end{array}$

Thus, the coefficient of term $x^n$ for $n=6$ is $-13608$ .