General term of binomial expansion — finding specific terms

WHY do we even need a "general term"?

WHAT we want: a formula that gives any term of $(a+b)^n$ by plugging in an index, so we can grab (say) the 7th term or the coefficient of $x^{10}$ without writing out all $n+1$ terms.

WHY it matters: expanding $(x+2)^{15}$ fully is 16 terms of pain. If a question only asks for "the coefficient of $x^9$ ", the general term lets us find it in one line.

Deriving it from scratch

$(a+b)^n = \underbrace{(a+b)(a+b)\cdots(a+b)}_{n \text{ brackets}}$

HOW a single term forms: to make one term of the product, you walk through all $n$ brackets and, from each, choose either $a$ or $b$ . Suppose you choose $b$ from exactly $r$ brackets. Then:

those $r$ brackets contribute $b^r$ ,
the remaining $n-r$ brackets contribute $a^{n-r}$ .

Why this step? Multiplication just picks one term from each bracket; the product is the product of your picks.

The number of different ways to choose which $r$ of the $n$ brackets give the $b$ is $==\binom{n}{r}==$ . All those ways give the identical term $a^{n-r}b^r$ , so they add up:

$\text{(all terms with } r \text{ b's)} = \binom{n}{r} a^{n-r} b^r$

Summing over every possible $r$ from $0$ to $n$ :

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

Figure — General term of binomial expansion — finding specific terms

The universal method (works every time)

To find a specific term or coefficient:

Write the general term $T_{r+1}=\binom{n}{r} a^{n-r} b^r$ with your actual $a,b,n$ .
Collect the power of the variable into one exponent.
Set that exponent equal to what the question demands → solve for $r$ .
Plug $r$ back to get the term/coefficient.

Worked Example 1 — a plain "which term"

Find the 5th term of $(2x+3)^7$ .

5th term means $r+1=5 \Rightarrow r=4$ . Why? Term number is $r+1$ .
Here $a=2x,\ b=3,\ n=7$ . $T_5 = \binom{7}{4}(2x)^{7-4}(3)^4 = 35\cdot(2x)^3\cdot 81$ Why this step? Direct substitution into the general term. $= 35 \cdot 8x^3 \cdot 81 = 22680\,x^3$

Worked Example 2 — coefficient of a specific power

Find the coefficient of $x^9$ in $(x^2 + \tfrac{1}{x})^{12}$ .

$a=x^2,\ b=x^{-1},\ n=12$ . $T_{r+1}=\binom{12}{r}(x^2)^{12-r}(x^{-1})^r = \binom{12}{r}\,x^{2(12-r)}\,x^{-r}$ Why? We separate the variable powers so we can combine them. $= \binom{12}{r}\,x^{24-2r-r} = \binom{12}{r}\,x^{24-3r}$ Set the exponent to 9: $24-3r = 9 \Rightarrow 3r = 15 \Rightarrow r = 5$ Why set equal to 9? We want the term containing $x^9$ . $\text{Coefficient} = \binom{12}{5} = 792$

Worked Example 3 — the term independent of $x$ (constant term)

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

$a=2x,\ b=-x^{-2},\ n=9$ . $T_{r+1}=\binom{9}{r}(2x)^{9-r}(-x^{-2})^r = \binom{9}{r}2^{9-r}(-1)^r\,x^{9-r}\,x^{-2r}$ $= \binom{9}{r}2^{9-r}(-1)^r\,x^{9-3r}$ "Independent of $x$ " means exponent $=0$ : $9-3r=0 \Rightarrow r=3$ Why 0? $x^0=1$ carries no $x$ . $T_4 = \binom{9}{3}2^{6}(-1)^3 = 84 \cdot 64 \cdot(-1) = -5376$

Common mistake Steel-man the classic slip-ups

Slip 1: "The 5th term uses $r=5$ ." Why it feels right: "5th → 5" is natural. The fix: $r$ starts at $0$ , so $T_{r+1}$ ; the 5th term is $r=4$ . Always write $r+1=$ term number first.

Slip 2: Forgetting the power on the whole first entry. Writing $(2x+3)^7$ 's term as $2\cdot x^3$ instead of $(2x)^3 = 8x^3$ . Why it feels right: you focused on $x$ . Fix: raise the entire bracket entry, coefficient and all.

Slip 3: Losing the sign in $\left(-\frac{1}{x^2}\right)^r$ . Why it feels right: you write $\frac{1}{x^{2r}}$ and drop the minus. Fix: factor as $(-1)^r x^{-2r}$ — the sign flips with parity of $r$ .

Slip 4: "Middle term is at $r=n/2$ always." Fix: even $n$ → single middle $T_{n/2+1}$ ; odd $n$ → two middle terms $T_{(n+1)/2}$ and $T_{(n+3)/2}$ .

Recall Feynman: explain it to a 12-year-old

Imagine you have $n$ boxes and each box hides either a red ball ( $a$ ) or a blue ball ( $b$ ). To build one "term," you open every box and read what's inside. If you decide "I want exactly $r$ blue balls," you multiply $r$ blues and $n-r$ reds together. But there are many ways to pick which boxes are blue — that count is $\binom{n}{r}$ . So each type of term appears $\binom{n}{r}$ times: that's exactly $\binom{n}{r}a^{n-r}b^r$ . To grab one special term, you don't open all boxes — you just say "how many blues make the power I need?" and jump straight there.

Connections

Binomial Theorem for positive integer index
Binomial coefficients and Pascal's Triangle
Middle term of a binomial expansion
Greatest coefficient and greatest term
Combinations nCr
Factorials

General term of

(a+b)^n

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

, with

r=0,1,\dots,n

Why is it

T_{r+1}

and not

T_r

Because

r

starts at

0

; the first term corresponds to

r=0

, so the

(r+1)

-th term uses index

r

Number of terms in

(a+b)^n

n+1

How to find coefficient of

x^k

Collect the variable's exponent in

T_{r+1}

, set it equal to

k

, solve for

r

, then evaluate the coefficient.

Term independent of

x

means what condition?

The net power of

x

equals

0

(x^2+1/x)^{12}

, exponent of

x

in general term

24-3r

Coefficient of

x^9

(x^2+1/x)^{12}

\binom{12}{5}=792

(from

r=5

Middle term rule

n

even → one middle term

T_{n/2+1}

;

n

odd → two middle terms

T_{(n+1)/2},T_{(n+3)/2}

5th term of

(2x+3)^7

r=4

T_5=\binom{7}{4}(2x)^3(3)^4=22680x^3

Common sign trap in

(-1/x^2)^r

Write as

(-1)^r x^{-2r}

; the sign depends on parity of

r

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab hum $(a+b)^n$ ko expand karte hain, to har term ek hi tarike se banta hai: $n$ brackets me se tumhe kuch me se $b$ lena hai aur baaki me se $a$ . Agar tum $r$ brackets me se $b$ choose karte ho, to $b^r$ banega aur bache $n-r$ brackets se $a^{n-r}$ . Aur "kaunse $r$ brackets me se $b$ lein" iske $\binom{n}{r}$ tarike hote hain — isliye general term ban jata hai $T_{r+1}=\binom{n}{r}a^{n-r}b^r$ . Isko yaad rakho: powers hamesha add hoke $n$ dete hain.

Sabse important baat — term number $r+1$ hota hai, $r$ nahi, kyunki $r$ shuru hota hai $0$ se. Isliye 5th term chahiye to $r=4$ lagao, seedha $r=5$ mat kar dena. Ye galti sabse common hai.

Specific term ya coefficient nikaalne ka trick simple hai: general term likho, variable ki saari power ko ek jagah collect karo, phir jo power question maang raha hai usko exponent ke barabar rakho, aur $r$ solve karo. Jaise "coefficient of $x^9$ " chahiye to exponent ko $9$ ke equal karo; "term independent of $x$ " chahiye to exponent ko $0$ ke equal karo. Yaad rakho jab $b$ me minus sign ho, jaise $(-1/x^2)^r$ , to usko $(-1)^r x^{-2r}$ likho — sign $r$ even/odd par depend karta hai, warna marks kat jaate hain!

General term of binomial expansion — finding specific terms

WHY do we even need a "general term"?

Deriving it from scratch

The universal method (works every time)

Worked Example 1 — a plain "which term"

Worked Example 2 — coefficient of a specific power

Worked Example 3 — the term independent of $x$ (constant term)

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Sequences & Series

Connections