Middle term and term independent of $x$

A binomial expansion $(a + b)^n$ has exactly $n + 1$ terms. Among these, one (or two) sits in the geometric centre , the middle term(s). Why care? Because the middle is typically where the binomial coefficient $\binom{n}{r}$ is largest, and a remarkable number of textbook and competitive problems hinge on isolating it. A close relative is the term independent of $x$, the constant lurking inside an expansion like $\left(x + \tfrac{1}{x^2}\right)^9$.

Definitions

In the expansion of $(a + b)^n$, $r$ runs from $0$ to $n$, so there are $n + 1$ terms numbered $T_1, T_2, \dots, T_{n+1}$.

• If $n$ is even, there is exactly one middle term: $T_{n/2 + 1}$.
• If $n$ is odd, there are two middle terms: $T_{(n+1)/2}$ and $T_{(n+3)/2}$.

A simple rule: count the number of terms, $n + 1$. If $n + 1$ is odd ($n$ even), the centre term is at position $(n+1+1)/2 = n/2 + 1$. If $n + 1$ is even ($n$ odd), the two centre terms straddle the middle.

The term independent of $x$ in an expansion that involves powers of $x$ is the term in which $x$ appears to the zeroth power , that is, the constant term. It need not be a middle term, and it may not exist at all if no admissible $r$ makes the exponent of $x$ vanish.

Why the middle is largest

For $0 \le r \le n$, the ratio of consecutive binomial coefficients is $\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n - r}{r + 1}.$ This ratio is greater than $1$ when $r < (n - 1)/2$ and less than $1$ when $r > (n - 1)/2$. So $\binom{n}{r}$ grows, hits a maximum near $r = n/2$, and decreases , the binomial coefficients are unimodal, peaking at the middle. This is one reason the middle term matters: it carries the largest coefficient.

Finding the middle term , recipe

1. Identify $n$. Decide whether it is even or odd.
2. Compute the middle index: $r = n/2$ if $n$ even, else $r = (n - 1)/2$ and $r = (n + 1)/2$.
3. Substitute into $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.
4. Simplify carefully , pay attention to signs and powers.

Worked examples

Example 1. Find the middle term in $(x + 2)^8$.

Here $n = 8$ is even, so there is one middle term at $r = 4$: $T_5 = \binom{8}{4} x^4 \cdot 2^4 = 70 \cdot 16 \cdot x^4 = 1120\, x^4.$

Example 2. Find the middle terms in $(2x - y)^7$.

$n = 7$ is odd. Middle terms are $T_4$ ($r = 3$) and $T_5$ ($r = 4$). $T_4 = \binom{7}{3} (2x)^4 (-y)^3 = 35 \cdot 16 x^4 \cdot (-y^3) = -560\, x^4 y^3.$ $T_5 = \binom{7}{4} (2x)^3 (-y)^4 = 35 \cdot 8 x^3 \cdot y^4 = 280\, x^3 y^4.$

Example 3. Find the middle term of $\left(x + \dfrac{1}{x}\right)^{10}$.

$n = 10$ even, middle at $r = 5$. $T_6 = \binom{10}{5} x^5 \cdot \frac{1}{x^5} = 252.$ The middle term is the constant $252$.

Example 4. Find the term independent of $x$ in $\left(\sqrt{x} - \dfrac{2}{x^2}\right)^{10}$.

$T_{r+1} = \binom{10}{r} (\sqrt{x})^{10-r}\left(-\dfrac{2}{x^2}\right)^r = \binom{10}{r}(-2)^r x^{(10-r)/2}\, x^{-2r} = \binom{10}{r}(-2)^r x^{(10 - r - 4r)/2} = \binom{10}{r}(-2)^r x^{(10 - 5r)/2}$.

Set $\dfrac{10 - 5r}{2} = 0 \Rightarrow r = 2$. So $T_3 = \binom{10}{2} (-2)^2 = 45 \cdot 4 = 180.$

Example 5. Find the middle term of $\left(\dfrac{x}{3} + 9y\right)^{10}$.

Middle term at $r = 5$: $T_6 = \binom{10}{5}\left(\frac{x}{3}\right)^5 (9y)^5 = 252 \cdot \frac{x^5}{3^5} \cdot 9^5 y^5 = 252 \cdot \frac{9^5}{3^5} x^5 y^5 = 252 \cdot 3^5 x^5 y^5 = 252 \cdot 243\, x^5 y^5 = 61\,236\, x^5 y^5.$

Try it yourself

1. Find the middle term of $(x + y)^6$.
2. Find the middle terms of $(2x - 3y)^7$.
3. Find the term independent of $x$ in $\left(x - \dfrac{1}{x}\right)^{12}$.
4. Find the term independent of $x$ in $\left(x^2 + \dfrac{1}{x^3}\right)^{15}$.
5. Find the middle term of $\left(\dfrac{x}{2} + \dfrac{2}{x}\right)^8$.
6. Show that the constant term of $\left(x + \tfrac{1}{x}\right)^{2n}$ is $\binom{2n}{n}$.
7. Find the middle term of $\left(1 + 3x + 3x^2 + x^3\right)^6$. (Hint: it is $(1 + x)^{18}$.)
8. In the expansion of $\left(x^2 - \dfrac{2}{x}\right)^9$, is there a term independent of $x$? Justify.
9. Find the coefficient of the middle term of $(1 + x)^{2n+1}$.
10. Show that the middle term of $(1 + x)^{2n}$ at $x = 1$ equals $\dfrac{(2n)!}{(n!)^2}$.
11. Find the ratio of the middle term to the second term in $(1 + x)^{10}$ at $x = 1$.
12. Compare the middle term of $(1 + x)^{12}$ with the sum of all other coefficients.

Pitfalls / Tricks

• For odd $n$ there are two middle terms , don't report only one.
• The term independent of $x$ is not necessarily the middle term.
• Always simplify the general term so that $x$ has a single exponent before setting it to zero.
• Insight. Among all $\binom{n}{r}$, the maximum is at (or just beside) $r = n/2$. The middle is where the action is.