3. General Term & Coefficients

Video coming soon3 worked examples

Theory

Often an exam asks only for one term — the term in $x^5$ , or the constant term — so expanding everything is wasted effort. Instead, use the general term (the $(r+1)$ th term) of $(a+b)^n$ :

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

The Golden Rule: find $r$ first. Write the general term, simplify the power of $x$ to a single expression in $r$ , set it equal to the power you need, and solve for $r$ . Only then substitute back to evaluate that one term.

⚠️ Common Examiner Traps

Off-by-one: it is the $(r+1)$ th term, so $r$ starts at $0$ . The term in $x^k$ is not the $k$ th term.
“The $k$ th term” means $r = k-1$ : a question asking for the fourth term needs $r = 3$ , not $r = 4$ .
“Independent of $x$ ” means power zero: the constant term is the one where the total power of $x$ equals $0$ .
Index algebra: when combining $x^{n-r}$ with $x^{-r}$ you get $x^{n-2r}$ — a very common place to slip.

Worked examples

Example 1

Find the term independent of $x$ in the expansion of $\left(2x + \dfrac{1}{x}\right)^6$ .

Step 1: Write the general term with $a = 2x$ , $b = \dfrac{1}{x}$ , $n = 6$ :

T_{r+1} = \binom{6}{r}(2x)^{6-r}\left(\frac{1}{x}\right)^r = \binom{6}{r}2^{6-r}\,x^{6-2r}

Step 2: “Independent of $x$ ” means the power of $x$ is zero:

6 - 2r = 0 \implies r = 3

Step 3: Substitute $r=3$ into the general term:

T_4 = \binom{6}{3}2^{3} = 20 \times 8 = 160

The term independent of $x$ is $160$ .

Example 2

Find the coefficient of $x^3$ in the expansion of $\left(x^2 - \dfrac{2}{x}\right)^6$ .

Step 1: Write the general term with $a = x^2$ , $b = -\dfrac{2}{x}$ , $n = 6$ :

T_{r+1} = \binom{6}{r}(x^2)^{6-r}\left(-\frac{2}{x}\right)^r = \binom{6}{r}(-2)^r\,x^{12-3r}

Step 2: We need the power of $x$ to be $3$ :

12 - 3r = 3 \implies r = 3

Step 3: Evaluate the coefficient at $r=3$ :

\binom{6}{3}(-2)^3 = 20 \times (-8) = -160

The coefficient of $x^3$ is $-160$ .

Example 3

Find the fourth term in the expansion of $(3a - 2b)^6$ , written in descending powers of $a$ .

Step 1: The general term is the $(r+1)$ th term, so the fourth term needs $r = 3$ :

T_{r+1} = \binom{6}{r}(3a)^{6-r}(-2b)^r \quad\implies\quad T_4 = \binom{6}{3}(3a)^{3}(-2b)^{3}

Step 2: Evaluate each factor separately, taking care to raise the whole bracket to the power:

\binom{6}{3} = 20, \qquad (3a)^3 = 27a^3, \qquad (-2b)^3 = -8b^3

Step 3: Multiply the three factors together:

T_4 = 20 \times 27a^3 \times (-8b^3) = -4320\,a^3b^3

The odd power of $-2b$ makes this term negative.