Question

Evaluate the value of \( (1.02)^8 \) using binomial theorem up to two decimal places.

• A. 1.17
• B. 1.18
• C. 1.81
• D. 1.71

Hint:

When using the binomial expansion for approximations, keep terms up to the desired accuracy and ignore higher-order terms if they contribute negligibly.

Answer 1

The Correct Option is A

Step 1: Apply the Binomial Theorem.
The binomial theorem states that:
\[ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots \]
Here, \( (1.02)^8 \) is of the form \( (1 + 0.02)^8 \), so we can use the binomial expansion for \( (1 + x)^n \), where \( x = 0.02 \) and \( n = 8 \).

Step 2: Substitute the values of \( x \) and \( n \).
We will expand \( (1 + 0.02)^8 \) using the binomial expansion and keep terms up to \( x^2 \) for better approximation.
\[ (1 + 0.02)^8 = 1 + 8 \times 0.02 + \frac{8(8-1)}{2!} \times (0.02)^2 + \cdots \]

Step 3: Calculate the terms.
- First term: \( 1 \)
- Second term: \( 8 \times 0.02 = 0.16 \)
- Third term: \( \frac{8 \times 7}{2!} \times (0.02)^2 = \frac{56}{2} \times 0.0004 = 0.0112 \)
So, the expansion becomes: \[ (1 + 0.02)^8 = 1 + 0.16 + 0.0112 + \text{higher order terms} \]

Step 4: Add up the terms.
Summing the terms up to \( x^2 \): \[ (1 + 0.02)^8 = 1 + 0.16 + 0.0112 = 1.1712 \]

Step 5: Round to two decimal places.
Rounding \( 1.1712 \) to two decimal places, we get \( 1.17 \).

Step 6: Conclusion.
The value of \( (1.02)^8 \) to two decimal places is \( 1.17 \), so the correct answer is option (A).