Question

use the Binomial Theorem to find the indicated coefficient or term.

The sixth term in the expansion of (3x +2)⁸

Short answer

The sixth term is

Step-by-step solution

8Step 1 . Given information  

Here (3x+2)⁸ is given

We have to find out the sixth term

. Expansion of  (3x+2)8 using The Binomial Theorem 

The binomial theorem states that

$$\begin{gathered} Letxandaberealnumbers.Foranypositiveintegern,wehave \\ {(x+a)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1}+\left(\begin{array}{c}n\\ 2\end{array}\right)a^2{x}^{n-2}+.......+\left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n-j}+.....+\left(\begin{array}{c}n\\ n\end{array}\right)a^n \\ thenbasedbonthistheorem{(3x+2)}^{8}becomes \\ (3x+2)8=\left(\begin{array}{c}8\\ 0\end{array}\right){\left(3x\right)}^8+\left(\begin{array}{c}8\\ 1\end{array}\right)\times {\left(2\right)}^1{\times \left(3x\right)}^{8-1}+\left(\begin{array}{c}8\\ 2\end{array}\right)\times {\left(2\right)}^2\times {\left(3x\right)}^{8-2}+\left(\begin{array}{c}8\\ 3\end{array}\right)\times {\left(2\right)}^3\times {\left(3x\right)}^{8-3}+\left(\begin{array}{c}8\\ 4\end{array}\right)\times {\left(2\right)}^4\times {\left(3x\right)}^{8-4}+\left(\begin{array}{c}8\\ 5\end{array}\right)\times {\left(2\right)}^5\times {\left(3x\right)}^{8-5}+.....+\left(\begin{array}{c}8\\ 8\end{array}\right)\times {\left(2\right)}^9 \\ =\left(\begin{array}{c}8\\ 0\end{array}\right){\left(3x\right)}^8+\left(\begin{array}{c}8\\ 1\end{array}\right)\times 2\times {\left(3x\right)}^7+\left(\begin{array}{c}8\\ 2\end{array}\right)\times 4\times {\left(3x\right)}^6+\left(\begin{array}{c}8\\ 3\end{array}\right)\times 8\times {\left(3x\right)}^5+\left(\begin{array}{c}8\\ 4\end{array}\right)\times 16\times {\left(3x\right)}^4+\left(\begin{array}{c}8\\ 5\end{array}\right)\times 32\times {\left(3x\right)}^3+....+\left(\begin{array}{c}8\\ 8\end{array}\right)256 \end{gathered}$$

Step 3.  Finding the sixth term 

Based on question , sixth term from the expansion is

$$\begin{gathered} \left(\begin{array}{c}8\\ 5\end{array}\right)\times 32\times {\left(3x\right)}^3=\frac{8!}{5!3!}\times 32\times 27\times x^3 \\ =\frac{8\times 7\times 6\times 5!}{3\times 2\times 1\times 5!}\times 32\times 27\times x^3 \\ =48384x^3 \end{gathered}$$