Question

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

The third term in the expansion of (3x - 2)⁹

Short answer

The third term is$314928x^7$

Step-by-step solution

. Given information  

In this question (3x-2)⁹ is given.

We have to find out the third term

. Expansion of  (3x-2)9 using The Binomial Theorem  

According to the binomial theorem

$$\begin{gathered} Letxandaberealnumbers.Foranypositiveintegern,wehave \\ {(x+a)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1}+\left(\begin{array}{c}n\\ 2\end{array}\right)a^2{x}^{n-2}+.......+\left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n-j}+.....+\left(\begin{array}{c}n\\ n\end{array}\right)a^n \end{gathered}$$

So given expression becomes

$$\begin{gathered} {(3x-2)}^9=\left(\begin{array}{c}9\\ 0\end{array}\right){\left(3x\right)}^9+\left(\begin{array}{c}9\\ 1\end{array}\right)\times {(-2)}^1{\times \left(3x\right)}^{9-1}+\left(\begin{array}{c}9\\ 2\end{array}\right)\times {(-2)}^2\times {\left(3x\right)}^{9-2}+.....+\left(\begin{array}{c}9\\ 8\end{array}\right)\times {(-2)}^8{\times \left(3x\right)}^{9-8}+\left(\begin{array}{c}9\\ 9\end{array}\right)\times {(-2)}^9 \\ =\left(\begin{array}{c}9\\ 0\end{array}\right){\left(3x\right)}^9-\left(\begin{array}{c}9\\ 1\end{array}\right)\times 2\times {\left(3x\right)}^8+\left(\begin{array}{c}9\\ 2\end{array}\right)\times 4\times {\left(3x\right)}^7+....+\left(\begin{array}{c}9\\ 8\end{array}\right)256\left(3x\right)-\left(\begin{array}{c}9\\ 9\end{array}\right)512 \end{gathered}$$

Step 3. Description of finding the  third  term

The third term from the expanded form of the given expression is

$$\begin{gathered} \left(\begin{array}{c}9\\ 2\end{array}\right)\times 4\times {\left(3x\right)}^7=\frac{9!}{2!7!}\times 4\times 3^7\times x^7 \\ =\frac{9\times 8\times 7!}{1\times 2\times 7!}\times 4\times 3^7\times x^7 \\ =314928x^7 \end{gathered}$$