Question 387756
Let {{{x = 150}}}
Let {{{E}}} = earnings at end of each year
At the end of the 1st year:
{{{E = x + .032x = 1.032x}}}
At the end of the 2nd year:
{{{E = x + .032x + .032*(x + .032x)}}}
{{{E = 1.032x + .032*1.032x}}}
{{{E = 1.032*1.032x}}}
{{{E = 1.032^2*x}}}
At the end of the 3rd year:
{{{E = 1.032^2*x + .032*1.032^2*x}}}
{{{E = 1.032*1.032^2*x}}}
{{{E = 1.032^3*x}}}
Substituting:
{{{1.032^3*150 = 1.0991*150}}}
{{{1.0991*150 = 164.866}}}
Interest earned = {{{164.866 - 150 = 14.87}}}
In 3 years $14.87 interst is earned