Binomial Estimation
Substitute a small value of x into a binomial expansion, and a handful of terms gives a fast, accurate estimate of a power that would otherwise need a calculator.
The idea
The expansion of $(1+x)^n$ is an identity — it's true for every value of $x$, not just algebraic ones. So if $x$ happens to be a small decimal, substituting it in turns the expansion into a numerical estimate of $(1+x)^n$, built entirely from arithmetic.
The reason this is useful — rather than just a curiosity — is that when $x$ is small, each successive power of $x$ is far smaller than the one before it. $x=0.01$ gives $x^2=0.0001$ and $x^3=0.000001$: each term of the expansion shrinks the estimate's error by roughly two decimal places. A handful of terms is usually enough to match a calculator to several decimal places.
$(1+x)^n = 1 + nx + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \cdots$ is exact. Truncating it — stopping after a few terms — introduces an error equal to everything left off. Because the coefficients $\binom{n}{r}$ grow only polynomially in $r$ while $x^r$ shrinks geometrically (for small $x$), the leftover terms rapidly become negligible. The smaller $x$ is, the fewer terms you need for a given accuracy.
Setting up an estimation
To estimate a number like $1.02^{10}$: recognise it as $(1+x)^n$ with $x=0.02$ and $n=10$, expand $(1+x)^{10}$ symbolically up to a convenient number of terms, then substitute $x=0.02$ and evaluate numerically.
Worked example 1
The true value of $1.01^8$ is $1.082856\ldots$, so this three-term estimate is accurate to about 4 decimal places already. The fourth term, $\binom{8}{3}(0.01)^3 = 56 \times 0.000001 = 0.000056$, shows exactly where the small remaining error comes from.
Worked example 2
The sign of $x$ alternates the signs of the terms in the expansion — but because $x^2$ is always positive regardless of the sign of $x$, the third term stays positive even when $x$ is negative. Substitute carefully and let the algebra handle the signs; don't try to guess them in advance.
Worked example 3
Here the bracket isn't $(1+x)^n$, so both terms carry powers — the substitution has to match both the "2" and the small part.
Forgetting that the leading "$64$" already carries all of $2^6$ leads to writing $1+192x+\ldots$ as if the bracket were $(1+x)^6$ — always check the constant term equals the full value of the non-$x$ part raised to the power $n$ before proceeding.
Worked example 4
Rather than guessing, compare each new term's size directly — once a term is smaller than the required accuracy, it (and everything after it) can be safely dropped.
Each successive term shrinks by roughly a factor of $x$ (adjusted slightly by the changing binomial coefficient), which is why a smaller $x$ needs fewer terms for the same accuracy, regardless of how large $n$ is. Comparing consecutive terms directly — rather than assuming a fixed number of terms is always "enough" — is the reliable way to justify how many terms a question needs.
Always double-check which value of $x$ makes the bracket equal the target number — writing $1.02^{10}$ as $(1+x)^{10}$ needs $x=0.02$, not $x=1.02$. Substituting the wrong value is the most frequent slip in this topic, and it's easy to catch by checking $1+x$ matches the number being estimated.
Check yourself
Use the first three terms of the expansion of $(1+x)^9$ to estimate $1.02^9$, giving your answer to 4 decimal places.
(Hint: expand $1+9x+\binom{9}{2}x^2$ and substitute x = 0.02.)