Can we find the coefficient of $x^2$ in $(5 + 4x)(1 - x/2)^{10}$?

Can we find the coefficient of $x^2$ in $(5 + 4x)(1 - x/2)^{10}$?

Question

Write down and simplify the first $4$ terms in the expansion by the binomial theorem of $\left(1 - \tfrac{1}{2}x\right)^{10}$.

Find the coefficient of $x^2$ in the expansion of $(5 + 4x)\left(1 - \tfrac{1}{2}x\right)^{10}.$

Solution

The

binomial theorem

says that, if

$x$

is a real number and

$n$

is a positive integer, then

$$(1+x)^n = \sum_{i=0}^n \binom{n}{i} x^i$$

where

$$\begin{align*} \binom{n}{0} &= 1, \\ \binom{n}{i} &= \frac{n(n-1) \dots m (n-(i-1))}{i(i-1) \dots m 1} \quad\text{for each integer }1 \le i \le n\text{.} \end{align*}$$

By applying this to the expression

$\left(1 - \tfrac{1}{2}x\right)^{10}$

, we have that

$$\left(1 - \tfrac{1}{2}x\right)^{10} = \binom{10}{0} \left( -\frac{x}{2} \right)^0 + \binom{10}{1} \left( -\frac{x}{2} \right)^1 + \binom{10}{2} \left( -\frac{x}{2} \right)^2 + \binom{10}{3} \left( -\frac{x}{2} \right)^3 + \dots b$$

We can simplify this a little.

$$\begin{align*} \binom{10}{1} &= \frac{10}{1} = 10, \\ \binom{10}{2} &= \frac{10 \times 9}{2 \times 1} = 45, \\ \binom{10}{3} &= \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120, \end{align*}$$

so that $\left(1 - \tfrac{1}{2}x\right)^{10} = 1 - 5x + \frac{45}{4} x^2 - 15x^3 + \dots b$

There are only two contributing terms to the coefficient of $x^2$.

The $5$ from the factor $(5 + 4x)$ pairs with the $\dfrac{45}{4}x^2$ term in the expansion of $\left(1 - \tfrac{1}{2}x\right)^{10}$, and the $4x$ term pairs with the $-5x$ term. Since we have already calculated the coefficients in this expansion this is a straightforward task.

Thus, we have that the coefficient of

$x^2$

in the product is

$$\begin{align*} 5 \times \frac{45}{4} + 4 \times (-5) &= \frac{225}{4} - 20 \\ &= \frac{145}{4}. \end{align*}$$