In short

The general term of (a + b)^n is T_{r+1} = \binom{n}{r}\,a^{n-r}\,b^r. The middle term is T_{(n/2)+1} when n is even (one middle term) or T_{(n+1)/2} and T_{(n+3)/2} when n is odd (two middle terms). A term independent of x is one where the power of x is zero. The greatest coefficient is \binom{n}{\lfloor n/2 \rfloor}. This article shows how to find each of these in any binomial expansion.

A JEE question asks: "In the expansion of \left(x^2 + \dfrac{1}{x}\right)^{12}, which term does not contain x?" Expanding all 13 terms by brute force is possible but painful. The general-term formula solves it in three lines: write the power of x in the general term, set it to zero, and solve for r.

That is the theme of this article. The binomial theorem gives you the expansion; the general term formula gives you surgical access to any single term without computing the rest.

The general term

From the binomial theorem, the expansion of (a + b)^n is:

(a + b)^n = \sum_{r=0}^{n} \binom{n}{r}\, a^{n-r}\, b^r

General term

The (r + 1)-th term of (a + b)^n is:

T_{r+1} = \binom{n}{r}\, a^{n-r}\, b^r \qquad (r = 0, 1, 2, \dots, n)

The expansion has n + 1 terms in total. The index r counts how many factors of b appear; the remaining n - r factors are a. Since r starts at 0, the first term is T_1 (with r = 0) and the last term is T_{n+1} (with r = n).

Finding a specific term. To find the k-th term, set r = k - 1 in the formula. The offset by 1 is the most common source of errors in binomial-expansion problems.

Indexing of terms in the binomial expansionA horizontal timeline showing terms T1 through T(n+1). T1 corresponds to r=0, T2 to r=1, and so on. The k-th term corresponds to r = k minus 1. Total number of terms is n + 1. (a + b)ⁿ — term indexing T₁ T₂ T₃ ··· Tₖ ··· T(n+1) r = 0 r = 1 r = 2 r = k−1 r = n total: n + 1 terms (r ranges from 0 to n)
The $(r+1)$-th term uses $r$ as the index. The first term has $r = 0$; the $k$-th term has $r = k - 1$; the last term has $r = n$.

Middle term(s)

The expansion of (a + b)^n has n + 1 terms. Whether there is one middle term or two depends on whether n is even or odd.

Case 1: n is even. There are n + 1 (odd) terms, so there is exactly one middle term. It is the \left(\frac{n}{2} + 1\right)-th term:

\text{Middle term} = T_{\frac{n}{2}+1} = \binom{n}{n/2}\, a^{n/2}\, b^{n/2}

For (a + b)^{10}: the middle term is T_6 = \binom{10}{5}\, a^5\, b^5 = 252\,a^5b^5.

Case 2: n is odd. There are n + 1 (even) terms, so there are two middle terms. They are the \left(\frac{n+1}{2}\right)-th and \left(\frac{n+3}{2}\right)-th terms:

T_{\frac{n+1}{2}} = \binom{n}{\frac{n-1}{2}}\, a^{\frac{n+1}{2}}\, b^{\frac{n-1}{2}} \qquad \text{and} \qquad T_{\frac{n+3}{2}} = \binom{n}{\frac{n+1}{2}}\, a^{\frac{n-1}{2}}\, b^{\frac{n+1}{2}}

For (a + b)^7: the two middle terms are T_4 = \binom{7}{3}\,a^4b^3 = 35\,a^4b^3 and T_5 = \binom{7}{4}\,a^3b^4 = 35\,a^3b^4.

Middle terms: one for even n, two for odd nTwo examples. Top: n=6 has 7 terms, with T4 highlighted as the single middle term. Bottom: n=7 has 8 terms, with T4 and T5 highlighted as the two middle terms. Middle terms n = 6: T₁ T₂ T₃ T₄ T₅ T₆ T₇ ← 7 terms, 1 middle r = 3: T₄ = C(6,3)a³b³ = 20a³b³ n = 7: T₁ T₂ T₃ T₄ T₅ T₆ T₇ T₈ ← 8 terms, 2 middle T₄ = C(7,3)a⁴b³ = 35a⁴b³ T₅ = C(7,4)a³b⁴ = 35a³b⁴ even n → 1 middle term | odd n → 2 middle terms
For $n = 6$ (even), the $7$ terms have a single middle term $T_4$. For $n = 7$ (odd), the $8$ terms have two middle terms $T_4$ and $T_5$. The middle term(s) always have the largest binomial coefficient.

Term independent of x

When the binomial involves powers of x in both a and b, the general term T_{r+1} has a power of x that depends on r. The term independent of x (also called the constant term) is the term where this power is zero.

Method. Write the power of x in T_{r+1} as a function of r. Set it equal to 0 and solve for r. If r is a non-negative integer \leq n, the corresponding term is independent of x. If not, there is no such term.

Example. For \left(x^2 + \dfrac{1}{x}\right)^{12}:

T_{r+1} = \binom{12}{r}(x^2)^{12-r}\left(\frac{1}{x}\right)^r = \binom{12}{r}\, x^{2(12-r)}\, x^{-r} = \binom{12}{r}\, x^{24 - 3r}

Set 24 - 3r = 0: r = 8. So the term independent of x is:

T_9 = \binom{12}{8} = \binom{12}{4} = 495
Finding the term independent of x in (x squared + 1/x) to the 12A diagram showing the power of x in the general term: 24 minus 3r. A number line from r=0 to r=12 shows the power of x decreasing. At r=8, the power is zero, giving the constant term T9 = C(12,8) = 495. (x² + 1/x)¹² — power of x in T(r+1) r=0 x²⁴ r=3 x¹⁵ r=6 x⁶ r=8 x⁰ r=9 x⁻³ r=12 x⁻¹² power of x = 24 − 3r at r = 8: power = 0, so T₉ = C(12,8) = 495 the constant term is 495
The power of $x$ in the general term is $24 - 3r$, which decreases linearly. It crosses zero at $r = 8$, giving the constant term $T_9 = \binom{12}{8} = 495$.

Greatest coefficient

The binomial coefficient \binom{n}{r} is largest when r is as close to n/2 as possible.

When n is even: the unique maximum is \binom{n}{n/2}.

When n is odd: there are two equal maxima: \binom{n}{(n-1)/2} = \binom{n}{(n+1)/2}.

To see why, examine the ratio of consecutive coefficients:

\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n - r + 1}{r}

This ratio is > 1 when r < (n+1)/2 (meaning \binom{n}{r} > \binom{n}{r-1}) and < 1 when r > (n+1)/2. The coefficients increase up to the centre and then decrease symmetrically.

When the expansion is (pa + qb)^n with specific numerical values of p and q, the greatest term (not just the greatest coefficient) depends on p, q, and the point of evaluation. The ratio test T_{r+1}/T_r > 1 identifies the crossover.

Binomial coefficients for n=8 forming a symmetric hill peaking at r=4A bar chart showing C(8,r) for r from 0 to 8. The bars increase from C(8,0)=1 to C(8,4)=70 and then decrease symmetrically back to C(8,8)=1. The peak at r=4 is highlighted in red. C(8, r) for r = 0, 1, ..., 8 0 1 2 3 4 5 6 7 8 1 8 28 56 70 56 28 8 1
For $n = 8$, the binomial coefficients form a symmetric hill. The greatest coefficient is $\binom{8}{4} = 70$ at the centre ($r = n/2 = 4$). The symmetry $\binom{8}{r} = \binom{8}{8-r}$ makes the hill a mirror image about $r = 4$.

Interactive: explore the general term

Drag the red point to choose r from 0 to 8 in the expansion of (x + 2)^8. The readouts show the term number, the binomial coefficient, and the numerical value of the term at x = 1.

Interactive: terms of (x+2) to the 8 at x equals 1A bar chart with r from 0 to 8 on the horizontal axis and term value at x=1 on the vertical axis. A draggable red point moves along the tops of the bars. Readouts show the term number, C(8,r), and the numerical value C(8,r) times 2 to the r. r T(r+1) 0 1 2 3 4 5 6 7 8 ↔ drag the red point
The terms of $(x + 2)^8$ evaluated at $x = 1$. Each bar is $\binom{8}{r} \cdot 1^{8-r} \cdot 2^r = \binom{8}{r} \cdot 2^r$. The greatest term is not at $r = 4$ (greatest coefficient) but shifts right because the factor $2^r$ grows with $r$.

Two worked examples

Example 1: Find the middle term of $\left(\dfrac{x}{3} + 9y\right)^{10}$

Step 1. Determine whether n is even or odd. Here n = 10 (even), so there is exactly one middle term.

Why: n + 1 = 11 terms, and the middle one is the 6th (position \frac{10}{2} + 1 = 6).

Step 2. Find the middle term's r-value. The 6th term has r = 5.

T_6 = \binom{10}{5}\left(\frac{x}{3}\right)^{10-5}(9y)^5

Why: T_{r+1} with r = 5 gives T_6. The power of a = x/3 is 5 and the power of b = 9y is 5.

Step 3. Compute the coefficient.

\binom{10}{5} = 252
\left(\frac{x}{3}\right)^5 = \frac{x^5}{243}
(9y)^5 = 9^5 \cdot y^5 = 59{,}049\, y^5

Why: 3^5 = 243 and 9^5 = (3^2)^5 = 3^{10} = 59{,}049.

Step 4. Multiply everything together.

T_6 = 252 \cdot \frac{x^5}{243} \cdot 59{,}049\, y^5 = 252 \cdot \frac{59{,}049}{243} \cdot x^5 y^5 = 252 \cdot 243 \cdot x^5 y^5 = 61{,}236\, x^5 y^5

Verify the numerical part: 59{,}049 / 243 = 243 (since 243 \times 243 = 59{,}049). Then 252 \times 243: 252 \times 200 = 50{,}400, 252 \times 43 = 10{,}836, total = 61{,}236.

Result. The middle term is 61{,}236\, x^5 y^5.

The 11 terms of (x/3 + 9y) to the 10 with T6 highlighted as the middle termEleven terms T1 through T11 shown in a row. T6 is highlighted in the centre. Below, the calculation: C(10,5) = 252, times (x/3) to the 5, times (9y) to the 5, equals 61236 x to the 5 y to the 5. (x/3 + 9y)¹⁰ — 11 terms T₁ T₂ T₃ T₄ T₅ T₆ T₇ T₈ T₉ T₁₀ T₁₁ middle term (r = 5) C(10,5) × (x/3)⁵ × (9y)⁵ 252 × x⁵/243 × 59049 y⁵ = 61,236 x⁵y⁵
The middle term of $(x/3 + 9y)^{10}$ is $T_6$, sitting at the exact centre of the $11$-term expansion. Its value $61{,}236\, x^5 y^5$ comes from three factors: the binomial coefficient $252$, the $a$-power $x^5/243$, and the $b$-power $59{,}049\, y^5$.

The large numerical coefficient (61{,}236) arises because 9y carries a factor of 9 that compounds rapidly when raised to the 5th power. In contrast, x/3 shrinks. The middle term captures the collision between these two trends.

Example 2: Find the term independent of $x$ in $\left(\sqrt{x} - \dfrac{3}{x^2}\right)^{10}$

Step 1. Identify a, b, n. Here a = \sqrt{x} = x^{1/2}, b = -3/x^2 = -3x^{-2}, n = 10.

Why: expressing everything in terms of powers of x makes it possible to track the exponent through the general term.

Step 2. Write the general term and compute the power of x.

T_{r+1} = \binom{10}{r}(x^{1/2})^{10-r}(-3x^{-2})^r = \binom{10}{r}(-3)^r\, x^{(10-r)/2}\, x^{-2r}
= \binom{10}{r}(-3)^r\, x^{(10-r)/2 - 2r} = \binom{10}{r}(-3)^r\, x^{(10-5r)/2}

Why: combine the exponents of x: \frac{10-r}{2} + (-2r) = \frac{10 - r - 4r}{2} = \frac{10 - 5r}{2}.

Step 3. Set the power of x to zero. \frac{10 - 5r}{2} = 0 \implies 10 - 5r = 0 \implies r = 2.

Why: the term independent of x is the term where the exponent of x is exactly 0. Solving gives r = 2, which is an integer in the range [0, 10], so such a term exists.

Step 4. Compute T_3.

T_3 = \binom{10}{2}(-3)^2\, x^0 = 45 \times 9 = 405

Spot-check the sign: (-3)^2 = 9 > 0, so the term is positive. And \binom{10}{2} = 45. The constant term is 405.

Result. The term independent of x is 405.

Power of x in the general term of (root x minus 3 over x squared) to the 10A graph showing the power of x as a function of r. The power is (10 minus 5r)/2. At r=0, power is 5. At r=2, power is 0 (highlighted). At r=4, power is minus 5. A horizontal line at power=0 intersects the graph at r=2. (√x − 3/x²)¹⁰ — power of x vs r 0 5 −5 r power 0 2 4 6 8 10 r = 2 → power = 0 T₃ = C(10,2) · (−3)² = 45 × 9 = 405
The power of $x$ in $T_{r+1}$ is $\frac{10 - 5r}{2}$, a decreasing linear function of $r$. It crosses zero at $r = 2$, marking the term independent of $x$. That term is $T_3 = 405$.

The linear relationship between the power of x and r is typical. Whenever a and b are both powers of x, the exponent in T_{r+1} is a linear function of r, and finding the independent term reduces to solving a linear equation.

Common confusions

Going deeper

If you can find specific terms using the general-term formula, identify the middle term(s), and locate terms independent of x, you are fully equipped for the standard binomial-expansion problems. The rest of this section explores a few extensions.

The greatest term — ratio method

To find which term T_{r+1} is greatest in absolute value, examine the ratio:

\left|\frac{T_{r+1}}{T_r}\right| = \frac{n - r + 1}{r} \cdot \left|\frac{b}{a}\right|

The terms increase (in absolute value) as long as this ratio exceeds 1:

\frac{n - r + 1}{r} \cdot \left|\frac{b}{a}\right| > 1 \quad \implies \quad r < \frac{(n+1)|b|}{|a| + |b|}

The greatest term is at r = \left\lfloor \frac{(n+1)|b|}{|a| + |b|} \right\rfloor, or at two consecutive values of r if the expression is exactly an integer.

Example. For (1 + 3)^8 (that is, a = 1, b = 3, n = 8): the critical r is \frac{9 \times 3}{1 + 3} = \frac{27}{4} = 6.75. So r = 6 gives the greatest term: T_7 = \binom{8}{6} \cdot 3^6 = 28 \times 729 = 20{,}412.

Multinomial middle terms

For expansions of the form (a + b + c)^n (the multinomial case), the concept of a "middle term" does not directly apply — the terms are indexed by pairs (r, s) rather than a single index. The greatest-coefficient problem generalises to finding the maximum of \frac{n!}{r!\,s!\,(n-r-s)!}, which occurs when r, s, and n - r - s are as equal as possible.

Connection to probability

The binomial coefficients \binom{n}{r} appear in the binomial probability distribution. If you flip a fair coin n times, the probability of getting exactly r heads is \binom{n}{r}/2^n. The "hill shape" of the coefficients — peaking at r = n/2 — explains why getting close to half heads is the most likely outcome. The greatest coefficient corresponds to the mode of the distribution.

Where this leads next