How to Find One Term of a Huge Expansion Without Writing the Rest

Expanding $(x+y{)}^2$ and $(x+y{)}^3$ by multiplication is quick, but $(x+y{)}^{10}$ has eleven terms and would fill a page. The binomial theorem writes every term at once and lets you pick out any single term, coefficient or approximation directly.

This lesson covers the theorem and its proof, Pascal's triangle, general and middle terms, and applications to coefficients and approximations.

**For every positive integer n, $(x+y{)}^n={\sum }_{r=0}^n{}^n{C}_r{x}^{n-r}{y}^r$, and the theorem is proved by mathematical induction, using Pascal's rule to combine coefficients.

The expansion:**

$$(x+y{)}^n={}^n{C}_0{x}^n+{}^n{C}_1{x}^{n-1}y+{}^n{C}_2{x}^{n-2}{y}^2+\cdots +{}^n{C}_n{y}^n$$

- There are $n+1$ terms
- Powers of x fall from n to 0 while powers of y rise from 0 to n
- The coefficients add up to $2^n$, found by putting $x=y=1$

Outline of the proof:

- Base case — for $n=1$, $(x+y{)}^1=x+y$
- Inductive step — assume the result for $n=k$, multiply both sides by $(x+y)$ and collect terms; each new coefficient is ${}^k{C}_r+{}^k{C}_{r-1}={}^{k+1}{C}_r$ by Pascal's rule, which is the result for $n=k+1$

Worked example. Expand $(2x+3{)}^3$:

$$(2x{)}^3+3(2x{)}^2(3)+3(2x)(3{)}^2+3^3=8x^3+36x^2+54x+27$$

Check at $x=1$: $5^3=125$ and $8+36+54+27=125$.

An everyday example. Compound interest on a bank deposit multiplies the amount by $(1+r{)}^n$, and the binomial expansion shows how the interest-on-interest terms build up.

The substance. The binomial coefficients depend only on n — the same numbers 1, 3, 3, 1 appear in every cube, whatever x and y are.

**In Pascal's triangle each row begins and ends with 1 and every inner number is the sum of the two numbers above it, so row n lists the coefficients ${}^n{C}_0,{}^n{C}_1,\dots ,{}^n{C}_n$ of $(x+y{)}^n$.

The first rows:

- Row 1: 1, 1
- Row 2: 1, 2, 1
- Row 3: 1, 3, 3, 1
- Row 4: 1, 4, 6, 4, 1
- Row 5: 1, 5, 10, 10, 5, 1

Why it works.** Adding two neighbours is exactly Pascal's rule, ${}^n{C}_{r-1}+{}^n{C}_r={}^{n+1}{C}_r$.

Worked example. Expand $(x-2{)}^5$ using row 5:

$$x^5-5x^4(2)+10x^3(4)-10x^2(8)+5x(16)-32=x^5-10x^4+40x^3-80x^2+80x-32$$

The signs alternate because the second term is negative. Check at $x=3$: $(3-2{)}^5=1$, and $243-810+1080-720+240-32=1$.

An everyday example. Tossing a coin 4 times gives 0, 1, 2, 3 or 4 heads in 1, 4, 6, 4 and 1 ways — row 4 of the triangle.

The substance. Every row is symmetric — a picture of the identity ${}^n{C}_r={}^n{C}_{n-r}$.

**The general term is $T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$; when n is even there is one middle term, $T_{\frac{n}{2}+1}$, and when n is odd there are two, $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$.

Worked example (general term).** Find the 4th term of ${\left(2x-\frac{1}{x}\right)}^7$. Here $r+1=4$, so $r=3$:

$$T_4={}^7{C}_3(2x{)}^4{\left(-\frac{1}{x}\right)}^3=35\times 16x^4\times \left(-\frac{1}{x^3}\right)=-560x$$

Worked example (middle term). In $(x+2{)}^8$, n is even, so the single middle term is $T_5$:

$$T_5={}^8{C}_4{x}^4{2}^4=70\times 16x^4=1120x^4$$

For odd n, such as $(a+b{)}^7$, the two middle terms are $T_4$ and $T_5$, with equal coefficients ${}^7{C}_3={}^7{C}_4=35$.

An everyday example. Guessing 8 true-or-false questions at random most often gives 4 correct answers, because the middle coefficient ${}^8{C}_4=70$ is the largest in its row.

The substance. **$T_{r+1}$ uses ${}^n{C}_r$, not ${}^n{C}_{r+1}$** — the term number is always one more than r.

**To find a specific term, write the general term, combine the powers of x into one exponent and set it equal to the required power; for approximations, write the number as $(1+h{)}^n$ with small h and keep only the first few terms.

Worked example (term independent of x).** In ${\left(x^2+\frac{1}{x}\right)}^9$:

$$T_{r+1}={}^9{C}_r(x^2{)}^{9-r}{\left(\frac{1}{x}\right)}^r={}^9{C}_r{x}^{18-3r}$$

Setting $18-3r=0$ gives $r=6$, so the term is ${}^9{C}_6=84$.

Worked example (coefficient). The coefficient of $x^5$ in $(1+x{)}^{10}$ is ${}^{10}{C}_5=252$.

Worked example (approximation). Find $(1.02{)}^8$ to three decimal places:

$$(1+0.02{)}^8=1+8(0.02)+28(0.02{)}^2+56(0.02{)}^3+\cdots \approx 1+0.16+0.0112+0.000448\approx 1.172$$

The exact value, $1.17166$ to five decimal places, also rounds to $1.172$.

An everyday example. Estimating how ₹1 lakh grows at 2 per cent for 8 periods gives about ₹1.17 lakh with this same approximation.

The substance. Approximation works only when h is small — the terms must shrink quickly for the discarded ones not to matter.