Question 204192: If there are 2 apples and every year the percentage of apples increase by 2%. How many apples are there in 6000 years?
For the first year the formula seems to work 2+1[2(.02)]=2.04
For the second year, I think it should be 2.04+1[2.04(.02)]=2.0808
because we have to take into consideration the initial increase and 2% of the initial increase for further increase.
For the third year, 2.0808+1[2.0808(.02)]=2.122416
And so on.
How do I go about calculating the problem without having to do it 6000 times?
Found 2 solutions by scott8148, Earlsdon:
Answer by scott8148(6628)   (Show Source):
Answer by Earlsdon(6294)  (Show Source):
You can put this solution on YOUR website! You can use the same formula one would use for a savings account in which you have deposited P dollars for t years at an interest rate of r.
 where A = the amount after t years.
Let's apply this to the growth of the number of apples rather than the number of dollars.
A = the number of apples after t years.
P = The starting number of apples or 2 apples.
r = The growth rate or 0.02 (2% expressed as a decimal)
t = 6000 years.
 Use your calculator for the exponentiation.
 Round to nearest whole number.
 or...
A = 8,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
...and that'll make a lot of apple pies
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