Question on binomial theorem: Find the term independent of $x$ in the expansion of $(2+x)\left(2x+\frac{1}{x}\right)^5$

Question

Find the term independent of $x$ in the expansion of $$(2+x)\left(2x+\frac{1}{x}\right)^5$$

I am only able to do this with one binomial, but when there are two I do not totally understand the process to do this.

Answer 1

First, rewrite the fifth power as a sum $$(2+x)\left(2x+\frac 1x\right)^5 = (2+x) \sum_{k=0}^5 {5\choose k} (2x)^\color{red}{5-k}\left(\frac 1x\right)^\color{blue}{k}$$ General idea: Note that to have a free (of $x$) term when multiplied by $(\color{orange}2+\color{purple}x)$, we must have a constant term (when multiplied by $\color{orange}2$ yields a free term) or term with $\frac{1}{x}$ (when multiplied by $\color{purple}x$ yields a free term) in the sum.

• Let's see if there is a constant term in the sum. To check this, we equate the red and blue powers since $x$'s should cancel each other: $$5-k = k \implies k = 2.5 \qquad \color{red}{\sf X}$$ So, there is no such term since $k$ was not an integer.

• Let's see if there is a term with $\frac{1}{x}$ in the sum. To check this, we set the $\color{red}{\text{red}} = \color{blue}{\text{blue}} - 1$ since after cancelling $x$'s, one $x$ should "survive" in the denominator: $$5-k = k - 1 \implies k = 3 \qquad \color{gree}{\checkmark}$$ Yes! There is such term when $k=3$: $${5\choose k} (2x)^{5-k}\left(\frac 1x\right)^k\Bigg|_{k=3} = {5\choose 3} \frac 4x$$

Finally, when we multiply the sum and $(2+x)$ our constant (free) term will be $$x \cdot {5\choose 3} \frac 4x = 4 {5\choose 3} = 40$$

Answer 2

You can make life a bit easier in first writing

$$(2+x)\left(2x+\frac 1x\right)^5 = \frac 1{x^5}\color{blue}{(2+x)(2x^2+1)^5}$$

Now, you see you need only calculate the coefficient of $x^5$ in $(2+x)(2x^2+1)^5$. This is easy since $(2x^2+1)^5$ contains only even powers of $x$ but you are looking for $x^5$. Hence using binomial expansion you get

\begin{eqnarray*}[x^5](2+x)(2x^2+1)^5 & = & [x^5](2+x)\binom 52(2x^2)^2 \\ & = & \binom 52 \cdot 4 = 40 \end{eqnarray*}