When a deposit is made for several years and rates are included to the capital, the rate is calculated with [color=#0000ff]compound interest[/color]:[br][br][math]\LARGE \textcolor{blue}{C(n)=C\cdot q^n},[/math][br][br]where [br] [table][tr][td][color=#0000ff]C [/color][/td][td]= capital[/td][/tr][tr][/tr][tr][td][color=#0000ff]C(n) [/color][/td][td]= increased capital[/td][/tr][tr][td][color=#0000ff]n[/color][/td][td]= number of interest time, e.g. years[/td][/tr][tr][td][color=#0000ff]q[/color][/td][td]= rate factor [math]\Large 1+\frac{i}{100}[/math][/td][/tr][/table]
A person deposits 2100€ for an account for 6 years. Interest rate is 2.1% p.a. How much money is in the account at the end, if no money is withdrawn during the deposit time?[br][br][table][tr][td]C [/td][td]= 2100€[/td][/tr][tr][td]n[/td][td]= 6[/td][/tr][tr][td]i [/td][td]= 2.1% p.a.[/td][/tr][tr][td]q[/td][td]= [math]\Large 1+\frac{2.1}{100}[/math]=1.021[/td][/tr][/table][br][math]\Large C(6) =2100€\cdot 1.021^6=2378.89€[/math]
If 1000 euros is invested at an annual interest rate of 6 %, compounded [br]semiannually, how long time will it take to have 2000 euros at that [br]account?[br][br]The interest rate in given as an annual interest rate but compounded [br]semiannually. So, the interest is paid twice a year to the amount [br]actually being at the account. The table below shows the actual money [br]after every payment:[br][br][br] [math]\Large\begin{eqnarray}[br]x_0&=&1000\\[br]x_1&=&x_0+\frac{1}{2}\cdot 0.06 x_0=1.03x_0\\[br]x_2&=&x_1+\frac{1}{2}\cdot 0.06 x_1=1.03x_1=1.03^2x_0\\[br]x_3&=&1.03^3x_0\\[br]\vdots[br]\\[br]x_n&=&1.03^nx_0\end{eqnarray}[/math][br][br][br][br]where [i]n[/i] = 2[i]t[/i]. Now, we have to only solve the equation[br][br][br] [math]\Large\begin{eqnarray}[br]&1.03^{2t}\cdot1000 &=&2000\\[br]\Leftrightarrow &2t\ln(1.03)&=&\ln(2)\\[br]\Leftrightarrow& t&\approx& 11.7[br]\end{eqnarray}[/math][br][br] [br]It takes almost 12 years to have 2000 euros at that account without any extra payments.