JEE Corner — Binomial Theorem for JEE Main and Advanced

The General Term: One Formula, Most of the Marks

The rationalised textbook chapter stops at full expansions; JEE asks for single terms — the 7th term, the coefficient of $x^5$, the term free of $x$. One formula answers them all:

Key Point (General Term): In the expansion of $(a + b)^n$,

$$T_{r+1} = {^nC_r}\,a^{n-r}\,b^{r}, \qquad r = 0, 1, \ldots, n$$

Note the offset: the term number is one more than $r$ — $T_1$ has $r = 0$, $T_7$ has $r = 6$.

The coefficient recipe: write $T_{r+1}$, collect the power of $x$ as a function of $r$, set it equal to the target power, solve for $r$, substitute back. For the term independent of $x$, set the power to zero.

Worked: in $\left(x^2 + \frac{1}{x}\right)^6$, $T_{r+1} = {^6C_r}x^{12-3r}$; zero power needs $r = 4$, giving $T_5 = {^6C_4} = 15$.

[JEE Tip] If solving for $r$ gives a non-integer or $r > n$, the requested term does not exist — that IS the answer (coefficient 0), and it is a favourite trap option.

Middle Terms and Coefficient Sums

Key Point (Middle Terms): $(a+b)^n$ has $n+1$ terms. If $n$ is even, there is one middle term, $T_{\frac{n}{2}+1}$. If $n$ is odd, there are two, $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$.

For $(1+x)^{2n}$ the middle term is $T_{n+1}$ with the greatest coefficient $^{2n}C_n$ of that row.

Sums of coefficients by substitution:

• All coefficients of a polynomial expansion $P(x)$: put $x = 1$. For $(1 + x - 3x^2)^{7}$: $(1 + 1 - 3)^7 = -1$.
• Alternating sum: put $x = -1$.
• Binomial coefficients only: $\sum {^nC_r} = 2^n$; even- and odd-position sums are $2^{n-1}$ each.

Term from the end: the $r$-th term from the end of $(a+b)^n$ is the $(n - r + 2)$-th from the beginning — or swap $a$ and $b$ and count from the front.

[JEE Tip] "Sum of coefficients" and "sum of binomial coefficients" differ: for $(2 + 3x)^5$ the first is $5^5$ (put $x=1$), the second is $2^5$. Read which one the question wants.

Solved Examples

Example 1: A specific term

Find the 4th term of $(x + 2)^9$.

Solution:

Step 1 — Convert term number to $r$. $T_4$ means $r = 3$ (the offset: $T_{r+1}$).

Step 2 — Write the term. $T_4 = {^9C_3}\,x^{9-3}\,2^3 = 84 \cdot x^6 \cdot 8$.

Step 3 — Evaluate. $672x^6$.

Takeaway: The term number is always one more than $r$ — convert first, then substitute.

Example 2: A two-variable coefficient

Find the coefficient of $x^6y^3$ in $(x + 2y)^9$.

Solution:

Step 1 — Locate the term. $y^3$ needs $r = 3$: $T_4 = {^9C_3}x^6(2y)^3$ — and the $x$-power $9 - 3 = 6$ matches ✓.

Step 2 — Evaluate. $84 \times 8 = 672$.

Takeaway: Match BOTH powers before evaluating — a mismatch means no such term.

Example 3: The 5th term with a coefficient base

Find the 5th term of $(a + 2b)^7$.

Solution:

Step 1 — Convert. $r = 4$: $T_5 = {^7C_4}a^{3}(2b)^4$.

Step 2 — Evaluate. $35 \times 16 = 560$: $T_5 = 560\,a^3b^4$.

Takeaway: $(2b)^4$ contributes $2^4 = 16$ — bracket the base before raising.

Example 4: Term independent of $x$

Find the term independent of $x$ in $\left(2x + \frac{1}{3x^2}\right)^9$.

Solution:

Step 1 — Write the general term and collapse the powers.

$$T_{r+1} = {^9C_r}(2x)^{9-r}\left(\frac{1}{3x^2}\right)^r = {^9C_r}\,2^{9-r}\,3^{-r}\,x^{9-3r}$$

Step 2 — Set the power to zero. $9 - 3r = 0$ gives $r = 3$.

Step 3 — Substitute back. $T_4 = {^9C_3}\,\frac{2^6}{3^3} = 84 \times \frac{64}{27} = \frac{5376}{27} = \frac{1792}{9}$.

Takeaway: Independent term = solve (power of $x$) $= 0$; keep the numeric factors exact as fractions.

Example 5: Classic — the 13th term is constant

Show the 13th term of $\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}$ is independent of $x$ and find it.

Solution:

Step 1 — Convert and write the term. $r = 12$: $T_{13} = {^{18}C_{12}}(9x)^6\left(-\frac{1}{3\sqrt{x}}\right)^{12}$.

Step 2 — Track the powers of $x$. $x^6$ from the first factor; $\left(x^{-1/2}\right)^{12} = x^{-6}$ from the second: net $x^0$ ✓.

Step 3 — Evaluate. The even power kills the minus sign: $^{18}C_{12} \cdot \frac{9^6}{3^{12}} = 18564 \cdot \frac{3^{12}}{3^{12}} = 18564$.

Takeaway: $9^6 = 3^{12}$ cancels the denominator exactly — engineered bases like this are a JEE signature.

Example 6: Coefficient of a negative power

Find the coefficient of $x^{-5}$ in $\left(x + \frac{1}{x^2}\right)^{10}$.

Solution:

Step 1 — Collapse the powers. $T_{r+1} = {^{10}C_r}\,x^{10-r}\,x^{-2r} = {^{10}C_r}\,x^{10-3r}$.

Step 2 — Solve. $10 - 3r = -5$ gives $r = 5$.

Step 3 — Evaluate. $^{10}C_5 = 252$.

Takeaway: Negative target powers work exactly like positive ones — solve the same linear equation in $r$.

Example 7: Single middle term ($n$ even)

Find the middle term of $\left(\frac{x}{3} + 9y\right)^{10}$.

Solution:

Step 1 — Locate the middle. $n = 10$ (even): one middle term, $T_{\frac{10}{2}+1} = T_6$, i.e. $r = 5$.

Step 2 — Write it. $T_6 = {^{10}C_5}\left(\frac{x}{3}\right)^5(9y)^5 = 252 \cdot \frac{9^5}{3^5}\,x^5y^5$.

Step 3 — Simplify the powers. $\frac{9^5}{3^5} = 3^5 = 243$: $T_6 = 252 \times 243\,x^5y^5 = 61236\,x^5y^5$.

Takeaway: Convert $9^5/3^5$ to a single power of 3 before multiplying — arithmetic stays small.

Example 8: Two middle terms ($n$ odd)

Find the middle terms of $\left(3 - \frac{x^3}{6}\right)^7$.

Solution:

Step 1 — Locate. $n = 7$ (odd): two middles, $T_4$ ($r = 3$) and $T_5$ ($r = 4$).

Step 2 — Compute $T_4$. $^7C_3\,3^4\left(-\frac{x^3}{6}\right)^3 = 35 \cdot 81 \cdot \left(-\frac{x^9}{216}\right) = -\frac{2835}{216}x^9 = -\frac{105}{8}x^9$.

Step 3 — Compute $T_5$. $^7C_4\,3^3\left(-\frac{x^3}{6}\right)^4 = 35 \cdot 27 \cdot \frac{x^{12}}{1296} = \frac{945}{1296}x^{12} = \frac{35}{48}x^{12}$.

Takeaway: Odd index → two middle terms with opposite signs here — the odd/even power of the negative part decides each sign.

Example 9: Middle term that collapses

Find the middle term of $\left(x - \frac{1}{x}\right)^8$.

Solution:

Step 1 — Locate. $n = 8$: middle is $T_5$, $r = 4$.

Step 2 — Write and simplify. $^8C_4\,x^4\left(-\frac{1}{x}\right)^4 = 70 \cdot x^4 \cdot x^{-4} = 70$.

Takeaway: Symmetric bases make the middle term the constant term — two questions, one answer.

Example 10: Sum of coefficients of a full expression

Find the sum of the coefficients of $(1 + x - 3x^2)^7$.

Solution:

Step 1 — Substitute $x = 1$. Sum of coefficients $= (1 + 1 - 3)^7$.

Step 2 — Evaluate. $(-1)^7 = -1$.

Takeaway: The substitution handles trinomials as easily as binomials.

Example 11: Two different "sums"

For $(2 + 3x)^5$, find (i) the sum of the coefficients (ii) the sum of the binomial coefficients.

Solution:

Step 1 — (i) All coefficients. Put $x = 1$: $(2+3)^5 = 5^5 = 3125$.

Step 2 — (ii) Binomial coefficients only. The $^5C_r$ alone: $2^5 = 32$.

Step 3 — Contrast. The first includes the powers of 2 and 3 baked into each term; the second ignores them.

Takeaway: Read the question — "coefficients" and "binomial coefficients" differ whenever the base isn't $(1+x)$.

Example 12: Equal coefficients locate $r$ (classic)

In $(1+x)^{34}$, the coefficients of the $(r-5)$-th and $(2r-1)$-th terms are equal. Find $r$.

Solution:

Step 1 — Convert term numbers to coefficient indices. The $k$-th term has coefficient $^{34}C_{k-1}$: here $^{34}C_{r-6}$ and $^{34}C_{2r-2}$.

Step 2 — Apply the equality rule. Either $r - 6 = 2r - 2$ (gives $r = -4$, rejected) or $(r-6) + (2r-2) = 34$.

Step 3 — Solve. $3r - 8 = 34$, so $r = 14$.

Takeaway: Always test both branches of $^nC_a = {^nC_b}$ — and reject the one that breaks the constraints.

Example 13: Coefficients in AP

The coefficients of the 2nd, 3rd and 4th terms of $(1+x)^n$ are in AP. Find $n$.

Solution:

Step 1 — Write the AP condition. $2\,{^nC_2} = {^nC_1} + {^nC_3}$.

Step 2 — Expand and simplify. $n(n-1) = n + \frac{n(n-1)(n-2)}{6}$; divide by $n$ and multiply by 6: $6(n-1) = 6 + (n-1)(n-2)$.

Step 3 — Solve the quadratic. $n^2 - 9n + 14 = 0$ gives $n = 7$ or $n = 2$.

Step 4 — Reject the impossible root. $n = 2$ has no 4th term. $n = 7$.

Takeaway: Solve the algebra, then audit the roots against the term-count constraint.

Example 14: Term from the end

Find the 7th term from the end of $(x + 2)^{12}$.

Solution:

Step 1 — Convert to a front count. The $r$-th term from the end is the $(n - r + 2)$-th from the start: $12 - 7 + 2 = 7$.

Step 2 — Compute $T_7$. $r = 6$: $^{12}C_6\,x^6\,2^6 = 924 \times 64\,x^6$.

Step 3 — Evaluate. $59136\,x^6$ — here the 7th from either end coincide (13 terms, 7 is the middle).

Takeaway: End-counted terms convert by $n - r + 2$ — or swap the two quantities and count from the front.

Example 15: Does the term exist?

Find the coefficient of $x^4$ in $\left(x^3 + \frac{1}{x^2}\right)^7$.

Solution:

Step 1 — Collapse the powers. $T_{r+1} = {^7C_r}\,x^{3(7-r)}\,x^{-2r} = {^7C_r}\,x^{21-5r}$.

Step 2 — Solve. $21 - 5r = 4$ gives $5r = 17$, $r = \frac{17}{5}$ — not an integer.

Step 3 — Conclude. The power $x^4$ never occurs: coefficient $0$.

Takeaway: A non-integer (or out-of-range) $r$ IS the answer — the term simply does not exist.