Find $n$ in this binomial expansion.

Question

Given $$(3-2x)\left(1+\frac{x}{2}\right)^n.$$

Find $n$ when the coefficient of $x$ is $7$. Hence find the coefficient of $x^2$.

Please show you working out with steps for me, this is a exam question I came across and need help with.

Answer 1

HINT: First use the binomial theorem to expand the power, and do a little algebra:

$$\begin{align*} (3-2x)\left(1+\frac{x}2\right)^n&=(3-2x)\sum_{k=0}^n\binom{n}k\left(\frac{x}2\right)^k\\ &=(3-2x)\sum_{k=0}^n\binom{n}k\left(\frac12\right)^kx^k\\ &=3\sum_{k=0}^n\binom{n}k\left(\frac12\right)^kx^k-2x\sum_{k=0}^n\binom{n}k\left(\frac12\right)^kx^k\\ &=3\sum_{k=0}^n\binom{n}k\left(\frac12\right)^kx^k-2\sum_{k=0}^n\binom{n}k\left(\frac12\right)^kx^{k+1}\;.\tag{1} \end{align*}$$

The $x$ term in the first summation of $(1)$ is the $k=1$ term; the $x$ term in the second summation is the $k=0$ term. The $x$ term when the expression is simplified to a polynomial in $x$ is therefore

$$3\binom{n}1\left(\frac12\right)^1x-2\binom{n}0\left(\frac12\right)^0x=\left(3\cdot n\cdot\frac12-2\cdot1\cdot1\right)x\;.\tag{2}$$

• You’re told that the $x$ term is actually $7x$. Combine this with $(2)$ to find $n$.
• Now use $(1)$ to find the coefficient of $x^2$.