Question

What is the approximate value of $(1.02)^8$ ?

• A. 1.171
• B. 1.175
• C. 1.177
• D. 1.179

Answer 1

Answer: (A)

$\left(1.02\right)^{8}=\left( 1 + 0.02\right)^{8} $
$ \left(1+x\right)^{n} = 1 + nx + \frac{n\left(n-1\right)}{2!} x^{2} + \frac{n\left(n-1\right)\left(n-2\right)}{3!} x^{3} + ..... n = 8, x= 0.02$
$ \left(1+0.02\right)^{8} = 1 +8 \times0.02 + \frac{8 \times7}{2!} .\left(0.02\right)^{2} + \frac{8.7.6}{3!} \left(.02\right)^{3} $
Neglecting higher terms
= 1 + 0.16 + 28 $\times $ 0.0004 + 56 $\times $ 0.000008
= 1 + 0.16 + 0.0112 = 1.171