A man deposits $5000.00 at the start of every year in to his bank account. the interest paid by the bank is 4% a year. any interest he receives is kept in the bank. Find the sum of his savings at the end of the 12th year (including interest received in the 12th year). if he withdraws a sum of $30,000.00 at the end of 12th year and he tops making further deposits, calculate his savings at the end of the 15th year. could you show how to do it in detail please !

Question

anonymous · Answer

1 st  Year  =   D

2 nd Year=    D+D(1+R)

3rd  Year =(  D+D(1+R))(1+R)+D =   D[(1+R)^2+(1+R)+1

Nth Year =     D(1+R)^n-1+........+(1+R)+ 1]

using geometric Sum

$$D\left(\frac{1-(1+R)^n}{-R}\right)$$

anonymous · Answer

$$5000\left(\frac{1-(1+.04)^{12}}{-.04}\right)$$=75129.

anonymous · Answer

Withdraw 30000
75129.-30000=45129

anonymous · Answer

Use a simple compound interest formula with 45129 as principle
45129(1+.04)^3