JEE Mains Maths · Binomial Theorem

Consecutive Coefficients and Special Terms

Questions about how neighbouring coefficients compare — their ratio, an A.P. or G.P. among them — and about particular terms: the middle term, the greatest term, and terms counted from the end.

All Binomial Theorem notes JEE Mains Maths strategy

Why this matters

Twenty-four PYQs. One identity does most of the work: the ratio of neighbouring binomial coefficients is a simple fraction in n and r, so a condition on three consecutive coefficients becomes two linear equations. The rest use symmetry to handle middle terms and terms from the end. Two ideas cover the page.

Concept 1 of 2: The ratio of neighbouring coefficients

Never expand the factorials. The ratio

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

turns 'three consecutive coefficients in the ratio

p:q:s

' into two linear equations in

n

and

r

. An A.P. condition

2\binom nr=\binom n{r-1}+\binom n{r+1}

becomes a quadratic after dividing by

\binom nr

. Pascal's rule

\binom nr+\binom n{r+1}=\binom{n+1}{r+1}

handles sums of neighbours.

Definition

$\frac{\binom nr}{\binom n{r-1}}=\frac{n-r+1}{r}$ .
$\frac{\binom{n+1}{r+1}}{\binom nr}=\frac{n+1}{r+1}$ , $\frac{\binom nr}{\binom{n-1}{r-1}}=\frac nr$ .
Pascal: $\binom nr+\binom n{r+1}=\binom{n+1}{r+1}$ .
A.P. of $\binom n{r-1},\binom nr,\binom n{r+1}$ : $(n-2r)^2=n+2$ .

Neighbour ratio

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

Worked example

Three consecutive coefficients of

(1+x)^n

are in the ratio

1:2:3

. Find

n

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 8 Apr 2023 · Q72Moderate

Example 1 · Binomial Theorem · Consecutive Coefficients and Special Terms

If the coefficients of the three consecutive terms in the expansion of

(1 + x)^{n}

are in the ratio

1:5:20

, then the coefficient of the fourth term is

Which r is which

Fix one labelling — say the middle coefficient is

\binom nr

— and write both ratios from it. Mixing

r

for the first term in one ratio and for the middle in the other gives equations with no integer solution.

Drill 14 more on the ratio of neighbouring coefficients

Concept 2 of 2: Middle, greatest and end terms

For even

n

there is one middle term,

T_{n/2+1}

, with the greatest binomial coefficient

\binom n{n/2}

; for odd

n

there are two. The

k

-th term from the end of

(a+b)^n

is the

k

-th term from the start of

(b+a)^n

, so the ratio of the

k

-th from the start to the

k

-th from the end is

\left(\frac ab\right)^{n-2k+2}

. The greatest term is where

\frac{T_{r+1}}{T_r}

drops below 1.

Definition

Middle term: $T_{n/2+1}$ ( $n$ even); $T_{(n+1)/2}$ and $T_{(n+3)/2}$ ( $n$ odd).
$k$ -th from the end $=T_{n-k+2}$ .
$\frac{T_k(\text{start})}{T_k(\text{end})}=\left(\frac ab\right)^{n-2k+2}$ .
$\frac{T_{r+1}}{T_r}=\frac{n-r+1}{r}\cdot\frac ba$ .

Start versus end

\frac{T_k\ \text{(from start)}}{T_k\ \text{(from end)}}=\left(\frac ab\right)^{n-2k+2}

Worked example

\left(2+\frac12\right)^{n}

, the 3rd term from the start is 16 times the 3rd term from the end. Find

n

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q62Moderate

Example 2 · Binomial Theorem · Consecutive Coefficients and Special Terms

In the expansion of

\left( \sqrt[3]{2}+\frac{1}{\sqrt[3]{3}} \right)^{n},n \in N

, if the ratio of

15^{\text{th~}}

term from the beginning to the

15^{\text{th~}}

term from the end is

\frac{1}{6}

, then the value of

\ ^{n}C_{3}

is:

The exponent is n - 2k + 2

Between the

k

-th term from the start and the

k

-th from the end, the powers of

a

differ by

n-2k+2

, not

n-2k

. Test with

k=1

: the first and last terms differ by

a^n

against

b^n

Drill 8 more on middle, greatest and end terms

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

The ratio of neighbouring coefficients
Neighbour ratio
$\frac{\binom nr}{\binom n{r-1}}=\frac{n-r+1}{r}$
Middle, greatest and end terms
Start versus end
$\frac{T_k\ \text{(from start)}}{T_k\ \text{(from end)}}=\left(\frac ab\right)^{n-2k+2}$

Watch out for (2)

Which r is which
→ The ratio of neighbouring coefficients
The exponent is n - 2k + 2
→ Middle, greatest and end terms

Test yourself on Binomial Theorem

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Binomial Theorem chapter test Drill all 24 questions