Evaluation of a decimal number raised to a power

Question

How can I evaluate (1.01)⁵ to 5 decimal places without using a calculator?

Answer 1

Use the binomial expansion for (1+x)ⁿ , which is given as :

(1+x)ⁿ = ∑ⁿ_m=0 ⁿC_m x^m = 1 + nx + n(n-1) x² / 2 + ..

where ⁿC_m = n! / {(n-m)!m!} is the number of ways in which one can choose a subset of m elements from a set of n elements.

The motivation for this formula is pretty simple:

1. Write (1+x)ⁿ as (1+x)·(1+x)·......n-times,
2. We can see that when we expand all the brackets, there will be 2ⁿ terms, as there are n brackets, and each one has 2 terms : 1 and x. Many of these terms will be identical, which we will collect together. For instance, when n = 2, (1+x)·(1+x) = 1·1 + 1·x + x·1 + x·x = 1 + 2x + x²
3. Each of these 2ⁿ terms will recieve exactly one contribution from each bracket: either 1 or x.
4. We can see that the x^m term comes about when m out of n brackets contribute x (and the rest contribute 1).
5. The number of ways this can happen is the coefficient of x^m in the binomial expansion, and this is exactly the number of ways in which we can choose m out of n brackets.

Here, x = 1/100 , n = 5. So, we use:

(1+x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵

Since we only need the answer to 5 decimal places, we do not need all these terms.

We can see that 5x = 5/100 ; 10x² = 10/100² ; 10x³ = 10/100³ need to be taken into account, as their contributions appear within the 5th decimal place.

However, the last two terms 5x⁴ = 5/100⁴ and x⁵ = 1/100⁵ are less than (1/10)⁵ , so they do not need to be taken into account if we only want the answer to the 5th decimal place

So, (1.01)⁵ ≅ 1 + 5/100 + 10/100² + 10/100³