When n is even, middle term is the (n/2 + 1)th term.

Find the Middle Term of (1 + x)²⁰

Question

Find the middle term in the expansion of $(1 + x)^{20}$ .

Solution — Step by Step

Here $n = 20$ , which is even. For an even $n$ , there is exactly one middle term, and it sits at position $\dfrac{n}{2} + 1$ .

So the middle term is the $\left(\dfrac{20}{2} + 1\right) = 11$ th term.

The general term of $(1 + x)^n$ is:

T_{r+1} = \binom{n}{r} \cdot x^r

We need $T_{11}$ , which means $r + 1 = 11$ , so $r = 10$ .

T_{11} = \binom{20}{10} \cdot x^{10}

Now compute $\binom{20}{10}$ :

\binom{20}{10} = \frac{20!}{10! \cdot 10!} = 184756

So the middle term is $\mathbf{184756 \cdot x^{10}}$ .

Why This Works

The expansion of $(1+x)^n$ has $n+1$ terms total. When $n$ is even, $n+1$ is odd, so there’s a clean central term — the $\left(\frac{n}{2}+1\right)$ th one.

Think of it symmetrically: the coefficients form Pascal’s triangle, which is symmetric about the centre. For $n = 20$ , we have 21 terms, and the 11th sits exactly at the centre. The coefficient $\binom{20}{10}$ is also the largest in the entire expansion — you can verify this since $\binom{20}{k}$ is maximised at $k = 10$ .

The formula $T_{r+1} = \binom{n}{r} x^r$ works because we’re choosing which $r$ of the 20 brackets contribute an $x$ (and the remaining $20 - r$ brackets contribute 1). For the middle term, exactly half the brackets contribute $x$ .

Alternative Method

For competition speed, memorise this directly: middle term of $(1+x)^{2m}$ is always $\binom{2m}{m} x^m$ . Here $2m = 20$ , so $m = 10$ , giving $\binom{20}{10} x^{10}$ instantly — no setup needed.

With $(a + b)^n$ in general, the middle term formula extends to $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ . Since our base is $(1+x)^{20}$ , we have $a = 1$ and $b = x$ , so $a^{n-r} = 1^{10} = 1$ drops out cleanly. That’s why the answer looks simpler than the general case.

Common Mistake

Many students write the middle term as the $\frac{n}{2}$ th term (i.e., the 10th term here) instead of the $\left(\frac{n}{2}+1\right)$ th. This gives $T_{10} = \binom{20}{9} x^9$ , which is wrong. The confusion happens because $r$ in $T_{r+1}$ starts from 0, not 1. Always count: $(1+x)^{20}$ has 21 terms, numbered $T_1$ through $T_{21}$ , and the middle one is $T_{11}$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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