1.08^3 = 1.2597
1.25971 * 1000 = 1,259.71
4. At the end of every year for 3 years, RM1000 will be invested in an account that offers 8% compounded annualy.Find the account amount at the end of the 3 years.
2 answers
for annual deposits, just work it out. At the end of year
1: 1000
2: 1000*1.08 + 1000
3: 1000*1.08^2 + 1000*1.08 + 1000
...
n: 1000(1.08^(n-1) + ... + 1)
= 1000(1.08^n - 1)/(1.08-1)
or, for an interest rate of r, the balance at the end of n years is
1000((1+r)^n - 1)/(r)
1: 1000
2: 1000*1.08 + 1000
3: 1000*1.08^2 + 1000*1.08 + 1000
...
n: 1000(1.08^(n-1) + ... + 1)
= 1000(1.08^n - 1)/(1.08-1)
or, for an interest rate of r, the balance at the end of n years is
1000((1+r)^n - 1)/(r)