Binomial theorem for rational indices (approximate values): Worked examples

This page is a drill through every kind of problem the rational-index binomial theorem can throw at you. We do not introduce new theory — see the parent note for the derivation. Here we make sure that no matter what the exam hands you, you have already seen its twin.

Before we compute, one reminder in plain words:

The scenario matrix

Every problem this topic can throw lives in exactly one row of this table. The last column tells you which worked example nails it.

Cell	What makes it different	The hidden danger	Example
A. Root, $x > 0$	base slightly above 1, fractional $n$	none — easiest case	Ex 1
B. Reciprocal, $x > 0$	negative $n$ , base above 1	signs alternate	Ex 2
*C. Base slightly below* 1 ( $x < 0$ )**	must set $x$ negative	forgetting the minus sign propagates	Ex 3
D. Base far from 1	naive $x$ would be $> 1$	series diverges unless you factor first	Ex 4
E. Two-term product / correction	keep $x^{2}$ term for accuracy	dropping it too early	Ex 5
F. Large integer-ish power, tiny $x$	first-order only is enough	over-computing	Ex 6
G. Word problem (physics/percentage)	you must build $x$ and $n$	wrong small quantity	Ex 7
H. Exam twist — solve for unknown	approximation used backwards	direction of inequality	Ex 8
I. Degenerate / limiting	$x = 0$ , or $x \to 1^{-}$	knowing when it breaks	Ex 9

How to read the figure below. It is a number line of the base value $1 + x$ . The amber dashed line at $1$ is the anchor ( $x = 0$ , where the approximation is exact). The shaded cyan band from $0$ to $2$ is the region where $∣ x ∣ < 1$ and the series is allowed to converge; the amber bands outside it are where the series diverges and you must factor first (Cell D). The four white dots are the actual bases from Examples 1, 2, 3 and 5 — notice they all huddle close to the anchor, which is why keeping two or three terms suffices for them, while Cell D's base ( $1004$ ) sits far off the right of this window and needs the factoring trick.

Figure — Binomial theorem for rational indices (approximate values)

Worked Examples

Ex 1 — Cell A: a root just above 1

Ex 2 — Cell B: a reciprocal, base above 1

Ex 3 — Cell C: base slightly below 1 (negative $x$ )

Ex 4 — Cell D: base far from 1 (must factor first)

Ex 5 — Cell E: the $x^{2}$ correction earns its keep

Ex 6 — Cell F: big power, first order is plenty

Ex 7 — Cell G: word problem, build $x$ and $n$ yourself

Worked example A spherical balloon's radius grows by

$2%$ . By roughly what percent does its volume grow? ( $V = \frac{4}{3} π r^{3}$ .) Forecast: volume goes like $r^{3}$ , and $3 \times 2% = 6%$ — guess ≈ $6%$ .

Step 1. New radius $= r (1 + 0.02)$ , so new volume $= \frac{4}{3} π r^{3} (1 + 0.02)^{3}$ . The ratio is $(1 + 0.02)^{3}$ with $x = 0.02$ , $n = 3$ . Why this step? The constant $\frac{4}{3} π r^{3}$ cancels; only the growth factor matters, and it is exactly a binomial with a nice $x$ .

Step 2. $(1 + 0.02)^{3} \approx 1 + 3 (0.02) + \frac{3 \cdot 2}{2} (0.02)^{2} = 1 + 0.06 + 0.0012 = 1.0612$ . Why this step? Keeping $x^{2}$ captures the small extra beyond the naive $6%$ .

Step 3. Growth $= 1.0612 - 1 = 0.0612 = 6.12%$ . Why this step? Subtract the " $1$ " (original volume) to read off fractional increase.

Verify: exact $(1.02)^{3} = 1.061208$ , i.e. $6.1208%$ . The linear estimate $6%$ is the textbook "for small changes, percent in volume ≈ $3 \times$ percent in radius". ✓

Ex 8 — Cell H: the exam twist (invert the series approximation)

Worked example For small

$x$ , $(1 + x)^{1/2} \approx 1 + \frac{1}{2} x - \frac{1}{8} x^{2}$ . Given $(1 + x)^{1/2} = 1.01$ , use the series itself to solve for $x$ (to 3 dp). Forecast: if $1 + x = 1.01$ then $x$ should be a little above $2 \times 0.01 = 0.02$ .

Step 1. Set the approximation equal to the target: $1 + \frac{1}{2} x - \frac{1}{8} x^{2} = 1.01$ , i.e. $\frac{1}{2} x - \frac{1}{8} x^{2} = 0.01$ . Why this step? The task is to invert the series, not undo the square root — so we work with the truncated polynomial the theorem gives us.

Step 2 (first pass). Drop the tiny $x^{2}$ term: $\frac{1}{2} x \approx 0.01 \Rightarrow x_{0} \approx 0.02$ . Why this step? For small $x$ the linear term dominates, giving a starting estimate to refine.

Step 3 (refine). Put $x_{0} = 0.02$ back into the $x^{2}$ term: $\frac{1}{2} x - \frac{1}{8} (0.02)^{2} = 0.01$ , so $\frac{1}{2} x = 0.01 + 0.00005 = 0.01005$ , giving $x = 0.0201$ . Why this step? One iteration accounts for the correction we dropped; this is exactly how "solve the series backwards" works in an exam.

Step 4 (round). The question asks for 3 dp, so $x = 0.0201$ rounds to $x = 0.020$ . Why this step? Always finish by matching the requested precision — the raw $0.0201$ becomes $0.020$ to 3 dp.

Verify: feed $x = 0.0201$ into the exact function: $(1 + 0.0201)^{1/2} = 1.010000 \dots$ ✓ — the series-inversion answer is correct.

Ex 9 — Cell I: degenerate and limiting cases

Worked example (a) What is

$(1 + 0)^{n}$ ? (b) What happens to $(1 - x)^{- 1} = 1 + x + x^{2} + \dots$ as $x \to 1^{-}$ ? Forecast: (a) obviously $1$ . (b) the geometric sum $\frac{1}{1 - x}$ blows up as $x \to 1$ .

Step 1 (a). Set $x = 0$ : every term with $x$ vanishes, leaving $1 + 0 + 0 + \dots = 1$ . And directly $(1 + 0)^{n} = 1^{n} = 1$ . The series is exact (all corrections zero). Why this step? $x = 0$ is the only input where the approximation has zero error — it's the anchor point the whole Maclaurin build sits on. See Maclaurin & Taylor Series.

Step 2 (b). $(1 - x)^{- 1} = 1 + x + x^{2} + \dots$ is a geometric series with ratio $x$ . Its sum is $\frac{1}{1 - x}$ only while $∣ x ∣ < 1$ . As $x \to 1^{-}$ , the sum $\to \frac{1}{1 - 1} = + \infty$ : the series diverges. Numerically at $x = 0.9$ , partial sums crawl toward $\frac{1}{0.1} = 10$ ; at $x = 0.99$ toward $100$ . Why this step? This is the boundary the parent's $∣ x ∣ < 1$ condition guards — cross it and no finite number of terms helps. See Convergence of Series.

Verify: $\frac{1}{1 - 0.9} = 10$ and $\frac{1}{1 - 0.99} = 100$ confirm the blow-up; $(1 + 0)^{n} = 1$ for any $n$ . ✓

Recall One-line strategy per cell

Each line below is written as situation ::: what to do — read the left of the triple-colon as the prompt and the right as the answer.

Above 1, root ::: set $x > 0$ , small, expand directly. Above 1, reciprocal ::: $n = - 1$ , signs alternate. Below 1 ::: set $x < 0$ ; watch the $x^{2}$ term keeps its sign. Far from 1 ::: factor out nearest perfect power first. Need accuracy ::: keep the $\frac{n ( n - 1 )}{2} x^{2}$ term. Word / percent ::: cancel constants, expand the growth factor, subtract 1.

Connections

Binomial theorem for rational indices (approximate values) (index 3.3.14) — the parent theory.
Infinite Geometric Series — Ex 9's limiting case is the GP sum $\frac{1}{1 - x}$ .
Convergence of Series — why $∣ x ∣ < 1$ is non-negotiable.
Error & Approximation in Calculus — truncation error ≈ first dropped term (Ex 5).
Maclaurin & Taylor Series — $x = 0$ is the anchor point (Ex 9a).