# Find an approximation of (0.99)⁵ using the first three terms of its expansion

**Solution:**

We can write 0.99 as 0.99 = 1 - 0.01.

Then we get (0.99)⁵ = (1 - 0.01)⁵.

We can easily expand this using the binomial theorem upto the first three terms.

(0.99)⁵ = (1 - 0.01)⁵

= ⁵C₀(1)⁵ - ⁵C₁ (1)⁴ (0.01) + ⁵C₂(1)³ (0.01)²

Here, we can calculate the binomial coefficients ⁵C₀, ⁵C₁, ... using the nCr formula.

= 1 - 5(0.01) + 10(0.0001)

= 0.951

NCERT Solutions Class 11 Maths Chapter 8 Exercise ME Question 7

## Find an approximation of (0.99)⁵ using the first three terms of its expansion

**Summary:**

The approximate value of (0.99)^{5} to be 0.951 using the first three terms of its expansion

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