4. Approximating Powers

Video coming soon3 worked examples

Theory

The Binomial Theorem gives a way of evaluating awkward powers such as $3.1^4$ or $1.02^6$ by hand. Split the number into a convenient part plus a small part, then expand:

3.1^4 = (3 + 0.1)^4, \qquad 1.02^6 = (1 + 0.02)^6, \qquad 0.97^5 = (1 - 0.03)^5

Because the small part is raised to ever higher powers, the terms shrink rapidly. If the index $n$ is a positive whole number the expansion terminates, so summing every term gives an exact answer. When a question asks only for a stated accuracy, you may stop once the remaining terms are too small to affect the last required figure.

The Golden Rule: choose the split so the second term is small and the first is easy to raise to powers — $(3 + 0.1)$ , not $(2 + 1.1)$ . Then keep expanding until the next term cannot change the digit you have been asked for, and state clearly why you stopped.

⚠️ Common Examiner Traps

Rounding too early: keep full accuracy in every term and round only at the very end, or the final digit will be wrong.
Stopping too soon: before truncating, check the size of the next term. Only discard it if it cannot affect the required decimal place.
Powers of the small part: $(0.1)^3 = 0.001$ , not $0.3$ — a very common slip when the arithmetic is done mentally.
Signs when subtracting: for $(1-a)^n$ the terms alternate in sign. Write the bracket as $(1 + (-a))^n$ to keep them straight.

Worked examples

Example 1

Use the Binomial Theorem to find the exact value of $2.2^4$ .

Step 1: Write the number as a binomial and expand with coefficients $1, 4, 6, 4, 1$ :

2.2^4 = (2 + 0.2)^4 = 2^4 + 4(2)^3(0.2) + 6(2)^2(0.2)^2 + 4(2)(0.2)^3 + (0.2)^4

Step 2: Evaluate each term:

\begin{aligned} 2^4 &= 16 \\ 4(8)(0.2) &= 6.4 \\ 6(4)(0.04) &= 0.96 \\ 4(2)(0.008) &= 0.064 \\ (0.2)^4 &= 0.0016 \end{aligned}

Step 3: Add them. The index is a positive whole number, so the expansion terminates and the total is exact:

2.2^4 = 16 + 6.4 + 0.96 + 0.064 + 0.0016 = 23.4256

Example 2

Use the Binomial Theorem to evaluate $1.02^6$ correct to 4 decimal places.

Step 1: Write $1.02^6 = (1 + 0.02)^6$ . Since the first term is $1$ , every power of it is $1$ and the expansion simplifies to:

(1+0.02)^6 = 1 + 6(0.02) + 15(0.02)^2 + 20(0.02)^3 + 15(0.02)^4 + \cdots

Step 2: Evaluate the terms in turn:

\begin{aligned} 6(0.02) &= 0.12 \\ 15(0.0004) &= 0.006 \\ 20(0.000008) &= 0.00016 \\ 15(0.00000016) &= 0.0000024 \end{aligned}

Step 3: Check the next term before stopping. It is $6(0.02)^5 = 0.0000000192$ , far too small to affect the fourth decimal place, so we may truncate here.

Step 4: Add the terms and round at the end:

1 + 0.12 + 0.006 + 0.00016 + 0.0000024 = 1.1261624\ldots \approx 1.1262

Example 3

Use the Binomial Theorem to evaluate $0.97^5$ correct to 5 decimal places.

Step 1: Write $0.97^5 = (1 - 0.03)^5$ . With a negative second term the signs alternate, using coefficients $1, 5, 10, 10, 5, 1$ :

(1-0.03)^5 = 1 - 5(0.03) + 10(0.03)^2 - 10(0.03)^3 + 5(0.03)^4 - (0.03)^5

Step 2: Evaluate each term, keeping full accuracy:

\begin{aligned} 5(0.03) &= 0.15 \\ 10(0.0009) &= 0.009 \\ 10(0.000027) &= 0.00027 \\ 5(0.00000081) &= 0.00000405 \\ (0.03)^5 &= 0.0000000243 \end{aligned}

Step 3: Combine with alternating signs:

1 - 0.15 + 0.009 - 0.00027 + 0.00000405 - 0.0000000243 = 0.8587340257\ldots

Step 4: Round to 5 decimal places: $0.97^5 \approx 0.85873$ .