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Compounding daily interest
- Thread starter hftransit
- Start date
1% interest is 0.01 of the "daily" amount.
After 1 day, the amount has grown to (1000)+0.01(1000)=1000(1.01)
1% interest is generated on this increased amount in the 2nd day.
After 2 days, the amount has grown to \(\displaystyle 1000(1.01)+0.01[(1000)(1.01)]=1000(1.01)^2\)
The pattern is....
the power corresponds to the number of elapsed days,
hence after 2600 days, the amount is destined to have grown to
\(\displaystyle 1000(1.01)^{2600}\ dollars\)
After 1 day, the amount has grown to (1000)+0.01(1000)=1000(1.01)
1% interest is generated on this increased amount in the 2nd day.
After 2 days, the amount has grown to \(\displaystyle 1000(1.01)+0.01[(1000)(1.01)]=1000(1.01)^2\)
The pattern is....
the power corresponds to the number of elapsed days,
hence after 2600 days, the amount is destined to have grown to
\(\displaystyle 1000(1.01)^{2600}\ dollars\)
chrisr said:1% interest is 0.01 of the "daily" amount.
After 1 day, the amount has grown to (1000)+0.01(1000)=1000(1.01)
1% interest is generated on this increased amount in the 2nd day.
After 2 days, the amount has grown to \(\displaystyle 1000(1.01)+0.01[(1000)(1.01)]=1000(1.01)^2\)
The pattern is....
the power corresponds to the number of elapsed days,
hence after 2600 days, the amount is destined to have grown to
\(\displaystyle 1000(1.01)^{2600}\ dollars\)
Chrisr: \(\displaystyle 1000(1.01)^{2600}\approx\)$172,017,184,400,000
hftransit said:Hello everybody,
I am wanting to know how much $1,000 will be after 2,600 days at 1% compounding interest per day - and can you please tell me the answer in both numbers and words?
Thank you very very kindly,
Harold
Hello everybody,
I am wanting to know how much $1,000 will be after 2,600 days at 1% compounding interest per day - and can you please tell me the answer in both numbers and words?
Thank you very very kindly,
Harold
Hi hftransit,
Using the compound interest formula:
\(\displaystyle A=P\left(1+\frac{r}{n}\right)^{nt}\)
t = the number of years so 2600 days is approx 2600/365 years.
n = the number of compounding periods in each year
\(\displaystyle A=1000\left(1+\frac{.01}{365}\right)^{365\cdot\frac{2600}{365}}\)
Comes out to about $1,074.
1% per year is \(\displaystyle \frac{1}{365}\)% per day.
Then, after 1 day, the amount has grown to
\(\displaystyle 1000+\frac{0.01}{365}(1000)=1000(1+\frac{0.01}{365})\) dollars
After 2 days, this has grown to
\(\displaystyle 1000(1+\frac{0.01}{365})+\frac{0.01}{365}(1000)(1+\frac{0.01}{365})\)
\(\displaystyle =1000(1+\frac{0.01}{365})^2\)
The pattern is..... number of days elapsed = index
Therefore after 2600 days,
the money ought to have increased to
\(\displaystyle 1000(1+\frac{0.01}{365})^{2600}\) dollars
Thanks masters!!
Then, after 1 day, the amount has grown to
\(\displaystyle 1000+\frac{0.01}{365}(1000)=1000(1+\frac{0.01}{365})\) dollars
After 2 days, this has grown to
\(\displaystyle 1000(1+\frac{0.01}{365})+\frac{0.01}{365}(1000)(1+\frac{0.01}{365})\)
\(\displaystyle =1000(1+\frac{0.01}{365})^2\)
The pattern is..... number of days elapsed = index
Therefore after 2600 days,
the money ought to have increased to
\(\displaystyle 1000(1+\frac{0.01}{365})^{2600}\) dollars
Thanks masters!!