Simple Applications

Approximating a Power Close to 1

When $x$ is small (say $|x| < 1$, and often much smaller), the terms of $(1+x)^n = {}^{n}C_0+{}^{n}C_1x+{}^{n}C_2x^2+\cdots$ shrink rapidly, because each successive term carries one more power of the small number $x$. Keeping only the first few terms -- and discarding the rest as negligible -- gives a fast, accurate way to compute a power such as $(1.02)^6$ or $(0.98)^{10}$ to several decimal places, without a calculator, by writing the base as $1+x$ for a small $x$ (positive or negative) and expanding only as far as the required accuracy demands.

Comparing Two Quantities

The same truncation idea, used the other way round, gives a quick way to decide which of two expressions is larger without computing either one exactly. Since every term of $(1+x)^n$ for $x>0$ is positive, dropping every term after the first two only ever decreases the total, so

This one inequality is often enough, by itself, to settle a "which is bigger" question: if $1+nx$ already exceeds the number being compared to, then $(1+x)^n$ certainly does too, since it is at least as large as $1+nx$.

Divisibility Results

The binomial theorem also gives clean proofs that an expression of the form $a^n - bn - 1$ (or similar) is always divisible by some fixed number, whenever $a=1+b$ for that same $b$. Writing $a^n=(1+b)^n$ and expanding,

so

and every single term of this remaining sum carries a factor of at least $b^2$ (since the smallest power appearing is $r=2$). Consequently $a^n-nb-1$ is always divisible by $b^2$, for every positive integer $n$ -- this is exactly the technique used in Example 3 (with $b=8$, divisibility by $64$) and Exercise: Applications, Q2 (with $b=10$, divisibility by $100$).

Two Further Standard Identities

Setting $a=b=1$ in the binomial theorem gives the sum of an entire row of Pascal's Triangle: …