Middle Terms in Binomial Expansion: Odd vs Even Powers

You see a question asking for the middle term of (2x + 3)^8. You remember something about dividing by two, but there is a small problem: do you put 4 or 5 into the formula?

That one-step difference is where many otherwise correct solutions go wrong.

The way out is to separate two jobs. First, find the position of the middle term. Then find the value of r needed to calculate it. Once those are separate, even large powers become manageable.

Let’s build the rule, use it in several different kinds of questions, and finish with a short practice set you can check yourself.

Why the power and the number of terms have opposite parity

For a positive integer n, the standard expansion of (a + b)^n has n + 1 term positions. The powers of b run from 0 through n, including both ends. This structure follows from NCERT’s Binomial Theorem chapter.

Think of arranging chairs in a row. Seven chairs have one central chair. Eight chairs have two central chairs, with the exact centre falling between them.

Now apply that idea to an expansion:

Power n	Number of terms n + 1	Middle position or positions
4	5	3rd
5	6	3rd and 4th
8	9	5th
9	10	5th and 6th
12	13	7th
13	14	7th and 8th

Notice what changes when you move from power 8 to power 9. There are now ten positions instead of nine, so the single central position becomes a central pair.

The formula you need, with the indexing explained

We will write C(n, r) for the binomial coefficient, read as “n choose r”:

C(n, r) = n! / [r!(n - r)!]

Here, ! means factorial. For example, 4! = 4 × 3 × 2 × 1, and 0! = 1.

The general term of (a + b)^n is:

T_(r+1) = C(n, r) × a^(n-r) × b^r

The notation T_5 means the fifth term, and a caret means a power: x^3 means x cubed. OpenStax’s binomial theorem lesson develops the coefficient formula and the method for finding a single term.

The first term uses r = 0. The second uses r = 1. So the fifth term uses r = 4, not r = 5.

For the kth term, always use r = k - 1.

You can also write the formula directly in terms of the position k:

T_k = C(n, k - 1) × a^(n-k+1) × b^(k-1)

Both forms say the same thing. Choose one and use it consistently.

Even power: find the single middle term

Let the power be n = 2m. There are 2m + 1 terms, with m positions on each side of the centre. The middle position is therefore m + 1.

For an even power n:

• Middle position: n/2 + 1.
• Value of r: n/2.
• Middle term: C(n, n/2) × a^(n/2) × b^(n/2).

For (a + b)^8, there are nine terms. Four are before the fifth term, and four are after it. Therefore, calculate T_5 using r = 4.

Example 1: Find the middle term of (2x + 3)^8

Here n = 8, a = 2x, and b = 3.

There are 9 terms, so the middle term is T_5 and r = 4.

T_5 = C(8, 4) × (2x)^4 × 3^4

= 70 × 16x^4 × 81

= 90,720x^4.

This is the complete term. If the question asked only for its coefficient, the answer would be 90,720.

A useful check: the two selected entries are each raised to the fourth power. That equal split is exactly what we expect in the central position of an even-power expansion.

Example 2: Find the middle term of (3p - 2q)^6

The subtraction must stay attached to the second entry:

a = 3p and b = -2q.

There are 7 terms. The middle term is T_4, so r = 3.

T_4 = C(6, 3) × (3p)^3 × (-2q)^3

= 20 × 27p^3 × (-8q^3)

= -4,320p^3q^3.

Although the outside power is even, this middle term is negative. Its sign comes from the odd power of the negative second entry.

Odd power: calculate both middle terms

Let n = 2m + 1. There are 2m + 2 terms. The centre falls between positions m + 1 and m + 2.

For an odd power n:

Quantity	First middle term	Second middle term
Position	(n + 1)/2	(n + 3)/2
Value of r	(n - 1)/2	(n + 1)/2

For instance, power 9 gives ten terms. The middle positions are 5 and 6, requiring r = 4 and r = 5.

Do not average the two terms unless the question explicitly asks for their average. “Find the middle terms” asks you to state both.

Example 3: Find the middle terms of (x + 2)^7

There are 8 terms. The middle positions are 4 and 5, so r = 3 and r = 4.

First middle term:

T_4 = C(7, 3) × x^4 × 2^3

= 35 × x^4 × 8

= 280x^4.

Second middle term:

T_5 = C(7, 4) × x^3 × 2^4

= 35 × x^3 × 16

= 560x^3.

The binomial coefficients, 35 and 35, are equal. The complete terms are different because their other factors differ.

If you need their sum, write 280x^4 + 560x^3, or factor it as 280x^3(x + 2). You cannot add the coefficients and call it 840x^4 because these are unlike terms.

Example 4: Find the middle terms of (2x - 1)^9

There are 10 terms. We need T_5 and T_6.

For T_5, r = 4:

T_5 = C(9, 4) × (2x)^5 × (-1)^4

= 126 × 32x^5 × 1

= 4,032x^5.

For T_6, r = 5:

T_6 = C(9, 5) × (2x)^4 × (-1)^5

= 126 × 16x^4 × (-1)

= -2,016x^4.

The adjacent terms have opposite signs. If you get both positive, check whether you lost the brackets around -1.

Fractions and reciprocals: keep the whole entry together

A fractional entry does not change the middle position. It only changes what you substitute into the general term.

Example 5: Find the middle term of (x/2 + 2/x)^8

Here x ≠ 0. The middle term is T_5 with r = 4.

T_5 = C(8, 4) × (x/2)^4 × (2/x)^4

= 70 × (x^4/16) × (16/x^4)

= 70.

The variable cancels completely. This middle term happens to be constant.

That does not mean every middle term is constant. The cancellation depends on the powers of x in the two entries.

Example 6: Find the middle term of (x^2 + 3/x)^6

Again, x ≠ 0. There are 7 terms, so calculate T_4 using r = 3.

T_4 = C(6, 3) × (x^2)^3 × (3/x)^3

= 20 × x^6 × 27/x^3

= 540x^3.

This middle term is not constant. To find the constant term, we would need a different calculation:

T_(r+1) = C(6, r) × x^(12-2r) × 3^r/x^r

= C(6, r) × 3^r × x^(12-3r).

A constant requires 12 - 3r = 0, giving r = 4. So the constant term is T_5:

C(6, 4) × 3^4 = 15 × 81 = 1,215.

Example 7: Two middle terms with unequal powers of x

Find the middle terms of (x^2 - 2/x)^5, where x ≠ 0.

There are 6 terms, so the middle positions are 3 and 4.

T_3 = C(5, 2) × (x^2)^3 × (-2/x)^2

= 10 × x^6 × 4/x^2

= 40x^4.

T_4 = C(5, 3) × (x^2)^2 × (-2/x)^3

= 10 × x^4 × (-8/x^3)

= -80x.

Do the numerical factors and variable powers separately. For the second answer, x^4/x^3 = x, not x^7. Division means subtracting exponents.

Can the two middle terms be equal?

They can, but equal central binomial coefficients alone are not enough.

For an odd power n = 2m + 1, the middle terms are:

T_(m+1) = C(2m + 1, m) × a^(m+1) × b^m

T_(m+2) = C(2m + 1, m + 1) × a^m × b^(m+1)

The coefficients are equal by the symmetry of combinations. When a and b are nonzero, dividing the second complete term by the first gives b/a. Therefore, those two terms are equal when a = b.

Example 8: Find x if the middle terms of (x + 3)^7 are equal

The two terms are:

T_4 = 35 × x^4 × 27 = 945x^4.

T_5 = 35 × x^3 × 81 = 2,835x^3.

Set them equal and bring everything to one side:

945x^4 - 2,835x^3 = 0.

945x^3(x - 3) = 0.

So x = 0 or x = 3.

At x = 0, both selected terms are zero. At x = 3, both have the same nonzero value. If the question specifies x > 0, only x = 3 remains.

This is why dividing by x^3 immediately can lose a valid solution. Check the domain and any nonzero condition before cancelling a variable factor.

Middle does not automatically mean greatest

The largest binomial coefficients lie in the centre of a row of Pascal’s triangle. But a complete term also contains powers of the two entries.

Consider this small expansion:

(1 + 10)^4 = 1 + 40 + 600 + 4,000 + 10,000.

The middle term is 600, in the third position. The largest term is 10,000, in the fifth position.

There is no contradiction. A question about the middle asks about location; a question about the greatest term asks about size.

Similarly, if the entries are expressions involving x, their numerical sizes depend on x. Do not declare a term greatest just because its position is central.

A five-line answer structure that prevents mistakes

For a written solution, keep these steps visible:

1. State the power n and the total number of terms n + 1.
2. Identify the middle position or positions.
3. Convert each position into r by subtracting 1.
4. Substitute the complete signed entries into the general-term formula.
5. Simplify coefficients, signs, and powers, then state the full term or terms.

For example, a compact solution to (p + 2q)^4 is:

“There are 5 terms, so the middle term is T_3, with r = 2. Hence T_3 = C(4, 2)p^2(2q)^2 = 24p^2q^2.”

That is enough working to show where the answer came from. You do not need to expand every term.

Common mistake	What to do instead
Calling n/2 the middle position for even n	Use n/2 + 1 for the position
Substituting r = 5 for T_5	Use r = 4
Giving only one answer when n is odd	Calculate both central positions
Using b = 2q in (3p - 2q)^6	Use b = -2q
Treating (2x)^4 as 2x^4	Raise both factors: 16x^4
Giving C(n, r) as the whole answer	Include both powered entries
Assuming the middle term is constant	Check the final variable exponent
Cancelling a variable that may be zero	Factor first and retain zero solutions

Try these questions before reading the answers

Write down the middle positions first, even if you can calculate the terms mentally.

1. Which positions are the middle terms of (a + b)^15?
2. Find the middle term of (x + 1)^10.
3. Find the middle terms of (x + 1)^5.
4. Find the middle term of (2x - 3)^4.
5. Find the middle terms of (x - 2)^5.
6. Find the middle term of (x/3 + 3/x)^6, where x ≠ 0.
7. Find the middle term of (x^2 + 1/x)^8, where x ≠ 0. Is it constant?
8. The middle terms of (x + 2)^5 are equal. Find all real values of x.

Answers with short working

1. T_8 and T_9. There are 16 terms. Use r = 7 and r = 8 if you need their values.

2. 252x^5. There are 11 terms, so T_6 = C(10, 5)x^5 = 252x^5.

3. 10x^3 and 10x^2. The six-term expansion has T_3 and T_4 in the centre.

4. 216x^2. T_3 = C(4, 2)(2x)^2(-3)^2 = 6 × 4x^2 × 9.

5. 40x^3 and -80x^2. Use r = 2 and r = 3. The powers of -2 are 4 and -8.

6. 20. T_4 = C(6, 3)(x/3)^3(3/x)^3 = 20. All variable and fractional factors cancel.

7. 70x^4, which is not constant. T_5 = C(8, 4)(x^2)^4(1/x)^4 = 70x^(8-4).

8. x = 0 or x = 2. The middle terms are 40x^3 and 80x^2. Their equality gives 40x^2(x - 2) = 0. Keep the zero solution unless the question excludes it.

Frequently asked questions

Why does an even power have only one middle term?

An even power n gives an odd number of positions, n + 1. An odd-length row has one central position, namely n/2 + 1.

Why does an odd power have two middle terms?

An odd power gives an even number of positions. The centre lies between positions (n + 1)/2 and (n + 3)/2, so both are middle terms.

What is the difference between T_r and T_(r+1)?

T_r means the rth term. The usual general-term formula is written for T_(r+1), with the second entry raised to r. For example, T_6 requires r = 5 in that formula.

Do I use the same rule for (a - b)^n?

Yes. The positions stay the same. Use -b as the second entry, so its sign is included when you raise it to the required power.

Are the two middle terms always equal?

No. For an odd power, their binomial coefficients are equal, but the complete terms usually differ. Powers and numerical factors must also be considered.

Is the middle term always the greatest term?

No. Middle describes position, while greatest describes numerical size. A large value of one entry can make a term near an end larger than the central term.

Is the middle term always independent of x?

No. It is independent of x only when its simplified exponent of x is zero. For example, the middle term of (x^2 + 3/x)^6 is 540x^3, so it still depends on x.

What if the question asks for the coefficient of the middle term?

First calculate the complete middle term. Then report the coefficient of its variable part. If the term is -4,320p^3q^3, its numerical coefficient is -4,320.

Does reversing the order of the entries change the middle terms?

Reversing a and b reverses the standard term order. The single middle term stays the same; when there are two middle terms, they exchange places. Keep each entry’s sign attached when reversing a subtraction.

Can I use these formulas for negative or fractional outside powers?

These middle-position formulas apply to finite binomial expansions with nonnegative integer powers. The usual infinite binomial series for a negative or non-integer power has no last term and therefore no central position. For a positive-integer school question, count the positions first and then substitute into the general term.