Question

If you have $ 1000 at 10 % interest compound yearly, over 1 over 30 years . Generalise your answer for yearly investment P0 with annual interest rate of q for total return after N-years.

Math physics problem...show the work with steps.

Answer 1

principal = $1000

rate = 10% = 0.1

time = 30 years

n = 1

amount = principal(1 +r/n)^nt

amount = 1000(1 + 0.1)^30 = 1000(1.1)^30

amount = $17449.402

2.

principal = Po
rate = q
time = 1
amount after 1 year = Po(1+q)
after 1st year
principal = Po + Po(1+q)
rate = q
time = 1
amount after 2 years = [Po + Po(1+q)](1+q)
                     = Po(1+q) + Po(1+q)^2

amount after n years = Po(1+q) + P0(1+q)^2+......+Po(1+q)^n
    the sequence is in gp
with a = P0(1+q) and r = 1+q
Sn = a(r^(n+1) -1)/r-1
Sn = Po(1+q)((1+q)^(n+1) - 1)/q
amount after n years = Po(1+q)((1+q)^(n+1) - 1)/q

Answer 2