Find the 5th term of the expansion of the binomial (3x-1) 7

Find the 5th term of the expansion of the binomial (3x-1)7

The 5th term =3^3 x binomial(7, 4) x^3 =945 x^3.

The 5th term will be given by :

C(7, 4) (3x)^3 * (1)^4  =    35 * 27 * x^3   =  945 x^3

Find the 5th term of the expansion of the binomial (3x-1)⁷

$\begin{array}{|rcll|} \hline (3x-1)^7 &=& \binom{7}{0}\cdot (3x)^7 + \binom{7}{1}\cdot (3x)^6\cdot(-1)^1 \\ &+& \binom{7}{2}\cdot (3x)^5\cdot(-1)^2 + \binom{7}{3}\cdot (3x)^4\cdot(-1)^3 \\ &+& \color{Maroon}\binom{7}{4}\cdot (3x)^3\cdot(-1)^4\color{black} + \binom{7}{5}\cdot (3x)^2\cdot(-1)^5 \\ &+& \binom{7}{6}\cdot (3x)^1\cdot(-1)^6 + \binom{7}{7}(-1)^7 \\ \hline \end{array}$

The 5th term of the binomial (3x-1)⁷  is:

$\begin{array}{|rcll|} \hline && \color{Maroon}\binom{7}{4}\cdot (3x)^3\cdot(-1)^4\color{black} \\ &=& \binom{7}{4}\cdot (3x)^3 \quad & | \quad \binom{7}{4} = \binom{7}{7-4} = \binom{7}{3} \\ &=& \binom{7}{3}\cdot (3x)^3 \\ &=& \binom{7}{3}\cdot 27x^3 \\ &=& \frac{7}{3}\cdot\frac{6}{2}\cdot\frac{5}{1}\cdot 27x^3 \\ &=& 7\cdot\frac{6}{6}\cdot 5\cdot 27x^3 \\ &=& 35\cdot 27x^3 \\ &\mathbf{=}& \mathbf{945x^3} \\ \hline \end{array}$