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)

${\left(1.02\right)}^8={\left(1+0.02\right)}^8$
${\left(1+x\right)}^n=1+nx+\frac{n\left(n-1\right)}{2!}{x}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{x}^3+.....n=8,x=0.02$
${\left(1+0.02\right)}^8=1+8\times 0.02+\frac{8\times 7}{2!}.{\left(0.02\right)}^2+\frac{\mathrm{8.7.6}}{3!}{\left(.02\right)}^3$
Neglecting higher terms
= 1 + 0.16 + 28 $\times$ 0.0004 + 56 $\times$ 0.000008
= 1 + 0.16 + 0.0112 = 1.171