In short

In the expansion of (a + b)^n with positive a, b, the terms first increase in absolute value and then decrease (or stay equal at most once). To find the greatest term, compute the ratio T_{r+2}/T_{r+1} and find the value of r at which the terms switch from increasing to decreasing. If \frac{(n-r)\,b}{(r+1)\,a} is exactly 1 for some integer r, two consecutive terms tie for the greatest.

A factory owner in Surat invests ₹10,000 at 8% annual compound interest. After 12 years, the amount is 10{,}000 \times (1.08)^{12}. The binomial theorem expands (1 + 0.08)^{12} into 13 terms. If you wanted a quick estimate of (1.08)^{12}, you would add the few largest terms and ignore the rest. But which terms are the largest? Is it the first term (1)? The second (12 \times 0.08 = 0.96)? The third (\binom{12}{2}(0.08)^2 = 0.4224)?

The terms rise, hit a peak, and then fall. The question "which term is the greatest?" is not just a textbook exercise — it tells you where most of the value of the expansion is concentrated.

The ratio test for consecutive terms

Take the expansion of (a + b)^n where a > 0 and b > 0. The (r+1)-th term is:

T_{r+1} = \binom{n}{r}\, a^{n-r}\, b^r

The next term is:

T_{r+2} = \binom{n}{r+1}\, a^{n-r-1}\, b^{r+1}

Divide the (r+2)-th term by the (r+1)-th:

\frac{T_{r+2}}{T_{r+1}} = \frac{\binom{n}{r+1}}{\binom{n}{r}} \cdot \frac{b}{a}

The ratio of consecutive binomial coefficients is:

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

So:

\frac{T_{r+2}}{T_{r+1}} = \frac{(n - r)\, b}{(r + 1)\, a}
The ratio of consecutive terms in a binomial expansionA diagram showing T sub r+1 and T sub r+2 side by side, with the ratio formula (n-r)b over (r+1)a between them. An annotation explains: when the ratio is greater than 1, terms are still increasing; when less than 1, terms are decreasing. Ratio of consecutive terms T(r+1) → (n−r)·b / (r+1)·a → ratio > 1 terms increasing ratio = 1 two terms tie ratio < 1 terms decreasing The ratio decreases as r increases (numerator shrinks, denominator grows)
The ratio $T_{r+2}/T_{r+1} = \frac{(n-r)b}{(r+1)a}$ tells you whether the terms are still growing. As $r$ increases, this ratio drops — so the terms eventually stop growing and start shrinking.

This ratio is a decreasing function of r: as r grows, the numerator (n - r)b shrinks and the denominator (r + 1)a grows. So the ratio starts above 1 (terms increasing), eventually crosses 1, and then stays below 1 (terms decreasing). The term just before the crossover — or the pair of terms at the crossover — is the greatest.

The method

Finding the greatest term in $(a + b)^n$ (with $a, b > 0$)

Set the ratio \frac{T_{r+2}}{T_{r+1}} \geq 1:

\frac{(n - r)\, b}{(r + 1)\, a} \geq 1

Solve for r:

r \leq \frac{(n+1)\,b}{a + b} - 1 = \frac{(n+1)\,b - a - b}{a + b} = \frac{nb - a}{a + b}

Let m = \frac{(n+1)\,b}{a+b}.

  • If m is not an integer, the greatest term is T_{r+1} where r = \lfloor m \rfloor (the floor of m). There is exactly one greatest term.
  • If m is an integer, then T_m = T_{m+1} — two consecutive terms are equal, and both are the greatest.

The quantity m = \frac{(n+1)b}{a+b} is the key number. Compute it, and the answer follows immediately.

Flowchart for finding the greatest termStep 1: compute m equals (n+1)b over (a+b). Step 2: check if m is an integer. If not, the greatest term is T with index floor of m plus 1. If yes, two terms tie: T sub m and T sub m+1. Finding the greatest term — decision flowchart Compute m = (n+1)b / (a+b) Is m an integer? No Yes Greatest term: T(⌊m⌋ + 1) Two greatest: T(m) and T(m+1) unique greatest term both terms are equal and largest
The decision is entirely determined by whether $m = \frac{(n+1)b}{a+b}$ is an integer or not. If not, there is one greatest term. If yes, two consecutive terms tie.

Why the terms rise and then fall

Think of it this way. In the expansion of (1 + x)^n with x > 0, the r-th term is \binom{n}{r}x^r. Moving from T_{r+1} to T_{r+2}, you multiply by \frac{(n-r)x}{r+1}. Early on, when r is small, n - r is large and r + 1 is small, so this multiplier is big — the terms grow. As r increases, n - r shrinks and r + 1 grows, so the multiplier drops below 1 and the terms start falling. The terms form a single peak — they never go up again after they start coming down.

Bar chart showing terms of (1+2) to the 8 rising then fallingEight bars showing the values of T1 through T9 in the expansion of (1+2) to the 8. The bars rise to a peak at T7 (the 7th term) and then fall. T7 has value C(8,6) times 2 to the 6 equals 1792. (1 + 2)⁸ = 3⁸ = 6561 — term values 0 1000 1500 2000 T₁=1 T₂=16 T₃=112 T₄=448 T₅=1120 T₆=1792 T₇=1792 T₈=1024 T₉=256
The nine terms of $(1 + 2)^8$. The terms rise to a peak at $T_6$ and $T_7$ (both equal to $1792$), then fall. Here $m = \frac{9 \times 2}{3} = 6$, an integer, so two terms tie for the greatest.

In this example, (1 + 2)^8 has m = \frac{9 \times 2}{1 + 2} = 6, which is an integer. So T_6 = T_7 = 1792, and both are the greatest term. You can verify: T_6 = \binom{8}{5} \cdot 1^3 \cdot 2^5 = 56 \cdot 32 = 1792, and T_7 = \binom{8}{6} \cdot 1^2 \cdot 2^6 = 28 \cdot 64 = 1792.

Interactive: watch the peak shift

Drag the red point to change x in (1 + x)^{10}. The bars show the value of each term \binom{10}{r}x^r. As x grows, the peak shifts to the right — later terms become dominant.

Interactive: greatest term of (1+x) to the 10A bar chart with r from 0 to 10 on the horizontal axis and term value on the vertical axis. A draggable point controls x. As x increases, the peak bar shifts rightward. r value 0 1 2 3 4 5 6 7 8 9 10 ↔ drag the red point
Drag to change $x$ and watch the peak of the term distribution shift. For small $x$, the first few terms dominate. For large $x$, the later terms are biggest.

Two worked examples

Example 1: Find the greatest term in the expansion of $(3 + 2x)^9$ when $x = 1$

Step 1. Substitute x = 1 to get numerical values. The expansion is (3 + 2)^9 = 5^9, with the general term T_{r+1} = \binom{9}{r} \cdot 3^{9-r} \cdot 2^r.

Why: the greatest term question is about the absolute numerical values, so you need to fix x first. Here a = 3 and b = 2.

Step 2. Compute m = \frac{(n+1)b}{a+b} = \frac{10 \times 2}{3 + 2} = \frac{20}{5} = 4.

Why: m is the key number that determines the answer. You need (n+1) = 10, b = 2, and a + b = 5.

Step 3. Since m = 4 is an integer, two consecutive terms tie: T_4 and T_5 are both the greatest.

Why: when m is an integer, the ratio T_{r+2}/T_{r+1} equals exactly 1 at r = m - 1 = 3, meaning T_4 = T_5 (using the ratio). So neither is strictly bigger than the other.

Step 4. Verify by computing both terms.

T_4 = \binom{9}{3} \cdot 3^6 \cdot 2^3 = 84 \times 729 \times 8 = 489{,}888

T_5 = \binom{9}{4} \cdot 3^5 \cdot 2^4 = 126 \times 243 \times 16 = 489{,}888

Both are equal — confirmed.

Result. The greatest term in (3 + 2)^9 is T_4 = T_5 = 489{,}888.

Terms of (3+2) to the 9 showing the two greatest terms T4 and T5A row of terms T1 through T10 with their numerical values. T4 and T5 are highlighted in red, both showing the value 489888. The formula m = 20/5 = 4 is shown above. (3 + 2)⁹ — finding the greatest term m = (9+1)·2 / (3+2) = 20/5 = 4 (integer → two terms tie) T₁ 19683 T₂ 118098 T₃ 314928 T₄ 489888 T₅ 489888 T₆ 391910 T₇ 208944 T₈ 69648 T₉–T₁₀ small Verification: T₄ = C(9,3) · 3⁶ · 2³ = 84 × 729 × 8 = 489,888 Verification: T₅ = C(9,4) · 3⁵ · 2⁴ = 126 × 243 × 16 = 489,888 Both T₄ and T₅ are the greatest — they are equal
In the expansion of $(3+2)^9$, the parameter $m = 4$ is an integer, so $T_4$ and $T_5$ are both equal to $489{,}888$ — the greatest value. The terms rise up to this pair and fall afterwards.

The integer value of m is the signal for a tie. If m had been, say, 4.3, then only T_5 (corresponding to r = \lfloor 4.3 \rfloor = 4) would have been the unique greatest term.

Example 2: Find the greatest term in the expansion of $(2 + 3x)^{10}$ when $x = \frac{2}{3}$

Step 1. Substitute x = \frac{2}{3}. The expansion becomes (2 + 3 \cdot \frac{2}{3})^{10} = (2 + 2)^{10} = 4^{10}, with a = 2 and b = 3x = 2.

Why: b is the second term of the binomial, which is 3x. Plugging in x = 2/3 makes b = 2. Now a = 2 and b = 2, so the two parts of the binomial are equal.

Step 2. Compute m = \frac{(n+1)b}{a+b} = \frac{11 \times 2}{2 + 2} = \frac{22}{4} = 5.5.

Why: m is not an integer, so there will be a unique greatest term.

Step 3. Since m = 5.5 is not an integer, \lfloor m \rfloor = 5, and the greatest term is T_{5+1} = T_6.

Why: the floor of m gives the value of r at the peak. The term index is r + 1 = 6.

Step 4. Compute T_6.

T_6 = \binom{10}{5} \cdot 2^5 \cdot 2^5 = 252 \times 32 \times 32 = 258{,}048

Result. The greatest term is T_6 = 258{,}048.

Greatest term of (2+2) to the 10 is T6A diagram showing m = 5.5, which is not an integer. Floor of 5.5 is 5, so the greatest term is T6. T6 is computed as C(10,5) times 2 to the 5 times 2 to the 5 = 258048. (2 + 2)¹⁰ — finding the greatest term m = 11 × 2 / 4 = 5.5 (not integer → unique greatest term) T₁ T₂ T₃ T₄ T₅ T₆ T₇ T₈ T₉ T₁₀ T₁₁ T₆ = C(10,5) · 2⁵ · 2⁵ = 252 × 1024 = 258,048 ⌊5.5⌋ = 5, so greatest term has r = 5 → T₆ Check: T₅ = C(10,4)·2⁶·2⁴ = 210×64×16 = 215,040 < 258,048 ✓
With $m = 5.5$ (not an integer), there is exactly one greatest term: $T_6 = 258{,}048$. You can verify that the neighboring terms $T_5 = 215{,}040$ and $T_7 = 215{,}040$ are both smaller.

Notice that a = b = 2 here, and both neighbors T_5 and T_7 are equal (215{,}040). When a = b, the distribution of terms is symmetric, and the greatest term sits right at the middle — which for even n is the middle term T_{n/2 + 1}, and for odd n the two middle terms both tie.

When the binomial has negative terms

If the expansion involves negative terms — say (a - b)^n with a, b > 0 — the terms alternate in sign. The concept of "greatest term" then refers to the greatest term in absolute value. Replace b with |b| everywhere in the formula, find the greatest |T_{r+1}|, and then attach the correct sign.

For (3 - 2)^9, the terms in absolute value are the same as those of (3 + 2)^9 — the same m = 4, the same two greatest terms. The only difference is that some terms carry a negative sign.

Common confusions

Going deeper

If you can compute m = \frac{(n+1)b}{a+b}, decide whether it is an integer, and read off the greatest term, you have the full technique. The ideas below are for readers who want to see the algebra behind the method and its connection to probability.

Proof that the terms form a single peak

The ratio \frac{T_{r+2}}{T_{r+1}} = \frac{(n-r)b}{(r+1)a} is a strictly decreasing function of r (for r = 0, 1, \ldots, n-1). At r = 0, the ratio is \frac{nb}{a}, which is at least 1 when b \geq a/n (true for most practical expansions). At r = n - 1, the ratio is \frac{b}{na}, which is less than 1 when a > b/n. Since the ratio decreases continuously and crosses 1 at most once, the sequence T_1, T_2, \ldots, T_{n+1} is unimodal — it increases, possibly has a flat top (two equal terms), and then decreases.

Connection to the mode of the binomial distribution

In probability, if you perform n independent trials each with success probability p = \frac{b}{a+b}, the probability of exactly r successes is \binom{n}{r} p^r (1-p)^{n-r}. The most probable number of successes — the mode — is the r that maximizes this expression. This is exactly the same optimization problem as finding the greatest term in (a + b)^n (after dividing by (a+b)^n). The formula m = (n+1)p for the mode of the binomial distribution is the probabilistic version of the m = \frac{(n+1)b}{a+b} formula derived here.

Greatest term for large n

When n is large, the greatest term is approximately at r \approx \frac{nb}{a+b}. The term values near this peak are well-approximated by a Gaussian (bell) curve — this is the content of the normal approximation to the binomial distribution, and it explains why the bar chart of terms looks increasingly bell-shaped as n grows.

Where this leads next