Question

Expand${\left(3x+1\right)}^4$using the Binomial Theorem

Short answer

The value of${\left(3x+1\right)}^4=81x^4+108x^3+54x^2+12x+1$

Step-by-step solution

Step 1. Defination

Let$x$and$a$be real numbers for any positive integer$n$, we have${\left(x+a\right)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1}+\cdots \cdots +\left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n-j}+\cdots \cdots +\left(\begin{array}{c}n\\ n\end{array}\right)a^n=\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)x^{n-j}{a}^j$

Step 2. Explanation

$\Rightarrow {\left(3x+1\right)}^4=\left(\begin{array}{c}4\\ 0\end{array}\right){\left(1\right)}^0{\left(3x\right)}^4+\left(\begin{array}{c}4\\ 1\end{array}\right){\left(1\right)}^1{\left(3x\right)}^3+\left(\begin{array}{c}4\\ 2\end{array}\right){\left(1\right)}^2{\left(3x\right)}^2+\left(\begin{array}{c}4\\ 3\end{array}\right){\left(1\right)}^3{\left(3x\right)}^1+\left(\begin{array}{c}4\\ 4\end{array}\right){\left(1\right)}^4{\left(3x\right)}^0$

$\Rightarrow {\left(3x+1\right)}^4=81x^4+108x^3+54x^2+12x+1$