Q.Find an approximation of $(0.99)^5$ using the first three terms of its expansion.

Concept: Binomial Theorem Expansion

We write $0.99 = 1 - 0.01$ and expand $(1 - x)^5$ using the binomial theorem, keeping only the first three terms.

Step 1:

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

Step 2:

Set $x = 0.01$. The first three terms are: …

The problem asks for an approximation of $(0.99)^5$ using the first three terms of its expansion. This is a classic application of the Binomial Theorem for a number slightly less than 1.

The core idea is to rewrite $0.99$ as $1 - 0.01$. Then $(0.99)^5 = (1 - 0.01)^5$. The binomial theorem tells us that for any real number $n$ and small $x$, we can expand $(1 + x)^n$ as a series. Here, $n = 5$ (a positive integer) and $x = -0.01$. Since $0.01$ is small, the terms get rapidly smaller, so just the first few terms give a good approximation.

Why does this work? Because each successive term in the expansion involves a higher power of $0.01$, which makes it much smaller than the previous term. For instance, $0.01^2 = 0.0001$, and $0.01^3 = 0.000001$. So dropping terms beyond the third introduces only a tiny error.

Let's work through it step by step.

1. Rewrite the base.

   $0.99 = 1 - 0.01$. So $(0.99)^5 = (1 - 0.01)^5$.

2. Apply the Binomial Theorem.

   For a positive integer $n$, the expansion is:

Here $n = 5$ and $x = -0.01$. So we substitute:

3. Compute the first three terms.

   The problem asks for the first three terms. That means we take terms up to $x^2$ (since the constant term is the first, the $x$ term is the second, and the $x^2$ term is the third).

   • First term: $1$
   • Second term: $5 \times (-0.01) = -0.05$
   • Third term: $\frac{5 \times 4}{2} \times (-0.01)^2 = 10 \times 0.0001 = 0.0010$

   So the sum of the first three terms is:

4. Check the magnitude of the next term (optional, for confidence). …