Binomial theorem applications — finding specific terms, coefficient patterns

Question

Find the coefficient of $x^5$ in the expansion of $(2x - 3)^8$ .

(CBSE 11 / JEE Main — Binomial Theorem)

Binomial Problem Type Decision Tree

flowchart TD
    A["Binomial Problem"] --> B{What is asked?}
    B -->|"Specific term (r+1)th"| C["T_{r+1} = nCr a^{n-r} b^r"]
    B -->|"Coefficient of x^k"| D["Set power of x = k, solve for r"]
    B -->|"Middle term"| E{"n even or odd?"}
    B -->|"Term independent of x"| F["Set power of x = 0"]
    B -->|"Sum of coefficients"| G["Put x = 1"]
    E -->|Even: n=2m| H["(m+1)th term"]
    E -->|Odd: n=2m+1| I["(m+1)th and (m+2)th terms"]

Solution — Step by Step

In the expansion of $(a + b)^n$ , the general $(r+1)$ th term is:

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

Here $a = 2x$ , $b = -3$ , $n = 8$ :

T_{r+1} = \binom{8}{r}(2x)^{8-r}(-3)^r = \binom{8}{r}2^{8-r}(-3)^r \cdot x^{8-r}

We need $x^{8-r} = x^5$ , so $8 - r = 5$ , giving $r = 3$ .

T_4 = \binom{8}{3}(2)^5(-3)^3

= 56 \times 32 \times (-27)

= 56 \times (-864)

\boxed{\text{Coefficient of } x^5 = -48384}

Why This Works

The binomial theorem expands $(a + b)^n$ into $n + 1$ terms. Each term has $a$ raised to a decreasing power and $b$ raised to an increasing power, with binomial coefficients $\binom{n}{r}$ in front. The power of $x$ in each term is determined by the exponent structure — setting this equal to the desired power gives us the specific term.

Alternative Method — Pascal’s Triangle for Small n

For small values of $n$ (up to 6-7), Pascal’s triangle gives the binomial coefficients quickly:

Row 8: 1, 8, 28, 56, 70, 56, 28, 8, 1

The $r = 3$ coefficient is 56 (4th entry, starting from 0). Then multiply by $2^5 \times (-3)^3$ as before.

For JEE Main, the “term independent of $x$ ” problem is a favourite. In expansions like $(x + 1/x)^n$ , the general term has $x^{n-2r}$ . Setting $n - 2r = 0$ gives the term independent of $x$ . This pattern appears every 2-3 years in JEE Main.

Common Mistake

The biggest error: forgetting the negative sign. In $(2x - 3)^8$ , the second term is $-3$ , not $3$ . So $(-3)^r$ alternates signs: positive for even $r$ , negative for odd $r$ . With $r = 3$ , $(-3)^3 = -27$ . Students who write $+27$ get the magnitude right but the sign wrong — costing full marks.

Another common slip: confusing $T_{r+1}$ with $T_r$ . The general term formula uses $r$ starting from 0 — so $T_4$ corresponds to $r = 3$ , not $r = 4$ .