- #1

- 459

- 5

- Thread starter Swapnil
- Start date

- #1

- 459

- 5

- #2

- 7

- 0

Well let us do this by using some figures.Swapnil said:

Lets say we have $100, and it increases in value at 5% per year.

[tex]C_1=100[/tex]

[tex]r=1.05[/tex]

Using

[tex]C = C_1\cdot {(r)}^n [/tex]

Therefore

[tex]C = 100\cdot {(1.05)}^n [/tex]

After one year it is

C=105

and so on...

Hopefully that helps

- #3

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

- #4

- 459

- 5

Say you have $100 dollars in a bank. It increases at the rate of 0.05 every year.

So at the end of the 1st year, the amount is:

[tex]A_1 = 100 + 0.05*100 = 105[/tex]

and at the end of the 2nd, 3rd, and 4th year the amount, respectively, is:

[tex]A_2 = 105 + 0.05*105 = 110.25 [/tex]

[tex]A_3 = 110.25 + 0.05*110.25 = 115.7625 [/tex]

[tex]A_4 = 115.7625 + 0.05*115.7625 = 121.550625[/tex]

My question is that how can we model this growth by the following formula:

[tex]A_n = A_0 (1+r)^n[/tex]

I mean, how did they derived such a formula?

- #5

- 367

- 1

[tex]A_n = (A_{n-2}*(1+r))*(1+r)=A_{n-2}*(1+r)^2[/tex]

and more generally,

[tex]A_n = A_{n-m}*(1+r)^m[/tex]

Plugging in [itex]n[/itex] for [itex]m[/itex] gives

[tex]A_n = A_{n-n}*(1+r)^n=A_0 *(1+r)^n[/tex]

- Last Post

- Replies
- 4

- Views
- 2K

- Replies
- 3

- Views
- 649

- Last Post

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 5K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 12

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 990