Question

Find the value of $(1.02{)}^6$, correct upto four places of decimals.

Answer 1

$(1.02{)}^6=(1+0.02{)}^6$

$={(1+\frac{2}{100})}^6$

$={}^6{C}_0{1}^6+{}^6{C}_1{1}^5\left(\frac{2}{100}\right)+{}^6{C}_6{\left(\frac{2}{100}\right)}^6$ [using Binomial theorem]

$=1+6.\frac{2}{100}+15.\frac{2^2}{(100{)}^2}+20.\frac{2^3}{(100{)}^3}+15.\frac{2^4}{(100{)}^4}+6.\frac{2^5}{(100{)}^5}+\frac{2^6}{(100{)}^6}$

$=1+0.12+0.0060+0.000160+0.00000249-0.0000000192+0.000000000064$

$=1.126162419264$.