i+i^1+i^2+.....+i^n

[cotatliad]

Is there a function for the following formula? i+i^1+i^2+.....+i^n

Thanks!

 

[RSXchin]

if you can explain it... yes.

 

[RSXchin]

will this work?

Function To_the_Power(ByVal Intger As Long, ByVal Ending_At As Long)
    For i = 1 To Ending_At
        To_the_Power = To_the_Power + (Intger ^ i)
    Next i
End Function

the first number is your base integer, and the second is how many iterations you want to do.

 

[cotatliad]

Trying to find out the nominal sum of a cash flow that grows by x% every year for a set number of years . so say i have 10 and it grows 10% a year for 2 years, it would be 10*(1.1^0+1.1^1+1.1^2)=33.1. i need to do this for much more complicated numbers. There is a equation for it but cannot remember right now and dont have enough time to figure it out. Thanks again.

 

[RSXchin]

if you need to do this with decimals, you need to use DOUBLE instead of long. and why "^0"? that just gives you 1. everytime.

here is your modified code

Function To_the_Power(ByVal Intger As Double, ByVal Ending_At As Long)
    For i = 0 To Ending_At
        To_the_Power = To_the_Power + (Intger ^ i)
    Next i
End Function

 

[pgc01]

Trying to find out the nominal sum of a cash flow that grows by x% every year for a set number of years . so say i have 10 and it grows 10% a year for 2 years, it would be 10*(1.1^0+1.1^1+1.1^2)=33.1. i need to do this for much more complicated numbers. There is a equation for it but cannot remember right now and dont have enough time to figure it out. Thanks again.

Hi

This means than in your first post you forgot the ^0 in the first operand and you want to know the value of

i^0+i^1+i^2+i^3...+i^n

The result is

(i^(n+1)-1)/(i-1)

That you can easily calculate with a formula

In your example, with the initial capital of 10, growing at a 10% rate for 2 years, you get

10*(1.1^0+1.1^1+1.1^2)

applying the formula, is equal to

10*(1.1^3-1)/(1.1-1)=10*(1.331-1)/0.1=10*0.331/.1=33.1

 

[RSXchin]

anything to the power of 0, is 1. infinity to the power of 0, is 1.

 

[pgc01]

anything to the power of 0, is 1.

... and 0 to the power of anything is?


Anyway, the i^0 which, as you say will be 1 for any growth rate, represents the initial capital. So the expression is equivalent to

1+i^1+i^2+i^3...+i^n

which is equal to, as I wrote:

(i^(n+1)-1)/(i-1)

that you can easily calculate with a formula

 

[cotatliad]

That was the formula I could not remember! Thank you very much!