- #1
Banaticus
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Homework Statement
If you deposit $deposit into a bank account each year and get y% annual interest, the simple formula is
\frac{deposit((1+y)^{years}-1)}{y}
How is this formula calculated? What's going on?
Let's look at $50, 6% interest, two years.
Maybe I'm supposed to be dropping $50 in just as the first year ends, so I don't build any interest on it the first year. I then build interest on it the second year, then drop in another $50 just as the year ends:
=(0*1.06+50)*1.06+50 no interest the first year + 50, then interest on that and another 50
=50*1.06+50
=103
That squares with the formula given earlier. So, let's look at three years
=(0*1.06+50)*1.06*1.06+50*1.06+50
That could be rewritten as
=50x^2+50x+50 where x is 1+interest and the variable is (years-1, years-2... 1)
That comes to the same answer as the original formula, $159.18, so things are working out.
Let's look at 7 years. How does:
50*1.06^6+50*1.06^5+50*1.06^4+50*1.06^3+50*1.06^2+50*1.06+50
get turned into:
\frac{50(1.06^7-1)}{.06}
Let's play around with it.
50(1.06^6+1.06^5+1.06^4+1.06^3+1.06^2+1.06+1)
So
1.06^6+1.06^5+1.06^4+1.06^3+1.06^2+1.06+1=\frac{1.06^7-1}{.06}
right? Maybe if I explicitly state the interest...
(1+.06)^6+(1+.06)^5+(1+.06)^4+(1+.06)^3+(1+.06)^2+(1+.06)+1
(1^6+.06^6)+(1^5+.06^5)+(1^4+.06^4)+(1^3+.06^3)+(1^2+.06^2)+(1+.06)+1
(1+.06^6)+(1+.06^5)+(1+.06^4)+(1+.06^3)+(1+.06^2)+(1+.06)+1
1+.06^6+1+.06^5+1+.06^4+1+.06^3+1+.06^2+1+.06+1
.06^6+.06^5+.06^4+.06^3+.06^2+.06+7
This isn't helping, it's not taking me in the direction that I want to go. How does:
50*1.06^6+50*1.06^5+50*1.06^4+50*1.06^3+50*1.06^2+50*1.06+50
which is 419.6918825 get turned into:
\frac{50(1.06^7-1)}{.06}
which is 419.6918825? How is that original equation calculated?
