Question

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

The coefficient of x³in the expansion of (2x+1)¹²

Short answer

The coefficient $x^{3}is$1760

Step-by-step solution

. Given information

here (2x+1)¹² is given

we need to find the coefficient of x³

 step 2 . Expansion of  (2x+1)12 using The Binomial Theorem

According to the binomial theorem

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

Then the expansion of (2x+1)¹²becomes

$$\begin{gathered} {(2x+1)}^{12}=\left(\begin{array}{c}12\\ 0\end{array}\right){\left(2x\right)}^{12}+\left(\begin{array}{c}12\\ 1\end{array}\right)\left(1\right){\left(2x\right)}^{12-1}+...........+\left(\begin{array}{c}12\\ 9\end{array}\right){\left(1\right)}^9{\left(2x\right)}^{12-9}+\left(\begin{array}{c}12\\ 10\end{array}\right){\left(1\right)}^{10}{\left(2x\right)}^{12-10}+\left(\begin{array}{c}12\\ 11\end{array}\right){\left(1\right)}^{11}{\left(2x\right)}^{12-11}+\left(\begin{array}{c}12\\ 12\end{array}\right)1^{12} \\ =\left(\begin{array}{c}12\\ 0\end{array}\right){\left(2x\right)}^{12}+\left(\begin{array}{c}12\\ 1\end{array}\right){\left(2x\right)}^{11}+.......+\left(\begin{array}{c}12\\ 9\end{array}\right){\left(2x\right)}^3+\left(\begin{array}{c}12\\ 10\end{array}\right){\left(2x\right)}^2+\left(\begin{array}{c}12\\ 11\end{array}\right){\left(2x\right)}^1+\left(\begin{array}{c}12\\ 12\end{array}\right) \end{gathered}$$

 Step 3. Description of finding the coefficient of x3

According to the question, the coefficient of x³ from the expanded term is

$$\begin{gathered} \left(\begin{array}{c}12\\ 9\end{array}\right){\left(2\right)}^3{\left(1\right)}^9,whichisequalto\frac{12!}{3!9!}\times 2^3\times 1=\frac{12\times 11\times 10\times 9!}{3\times 2\times 1\times 9!}\times 8 \\ =1760,thisiscoe.of{x}^3 \end{gathered}$$