How to make an exponential series?

>>a2=a2-(49*150*1.05^1)<<

I'm assuming this is a typo and that cell A2 contains:

=A1-(49*150*1.05^1)

Change this to:

=A1-(49*150*1.05^(ROW()-1))

and drag down.

--

Vasant





I assume that the refernce in cell A2 is to A1 and not to A2.
"sirsoto" <[email protected]> wrote in message
news:[email protected]...
>I want to make the following computation in the example cells below;
> Highlight the three cells where I entered my formula; Drag down until
> 300,000 equals zero (or near zero).
>
> a1=300,000-(49*150*1.05^0)
> a2=a2-(49*150*1.05^1)
> a3=a2-(49*150*1.05^2)
>
> The problem i am having is showing a series of exponential increases. I
> need 1.05 (which is 5%) to increase each year. Please advise. Sirsoto

> ... Drag down until
> 300,000 equals zero (or near zero).

Hi. not sure if this would be helpful, but the number of periods to get to
zero could be given as an equation. For this, there is no integer solution.

s=300000 'starting amount
b=49*150
r=.05

?Log(1+(r*s)/b)/Log(r+1)
22.7940616057939

If you wanted, you could put this in A1, and copy down. However, the
equation doesn't look similar to the original.
=447000 - 147000*1.05^ROW()

This comes from the equation for the value at any given period. Period 22
would be:
n=22
?(b - b*(1 + r)^n + r*s)/r
16986.6741715063

n=23
?(b - b*(1 + r)^n + r*s)/r
-4513.9921199184

--
Dana DeLouis
Win XP & Office 2003


"sirsoto" <[email protected]> wrote in message
news:[email protected]...
>I want to make the following computation in the example cells below;
> Highlight the three cells where I entered my formula; Drag down until
> 300,000 equals zero (or near zero).
>
> a1=300,000-(49*150*1.05^0)
> a2=a2-(49*150*1.05^1)
> a3=a2-(49*150*1.05^2)
>
> The problem i am having is showing a series of exponential increases. I
> need 1.05 (which is 5%) to increase each year. Please advise. Sirsoto

>>a2=a2-(49*150*1.05^1)<<

I'm assuming this is a typo and that cell A2 contains:

=A1-(49*150*1.05^1)

Change this to:

=A1-(49*150*1.05^(ROW()-1))

and drag down.

--

Vasant





I assume that the refernce in cell A2 is to A1 and not to A2.
"sirsoto" <[email protected]> wrote in message
news:[email protected]...
>I want to make the following computation in the example cells below;
> Highlight the three cells where I entered my formula; Drag down until
> 300,000 equals zero (or near zero).
>
> a1=300,000-(49*150*1.05^0)
> a2=a2-(49*150*1.05^1)
> a3=a2-(49*150*1.05^2)
>
> The problem i am having is showing a series of exponential increases. I
> need 1.05 (which is 5%) to increase each year. Please advise. Sirsoto

> ... Drag down until
> 300,000 equals zero (or near zero).

Hi. not sure if this would be helpful, but the number of periods to get to
zero could be given as an equation. For this, there is no integer solution.

s=300000 'starting amount
b=49*150
r=.05

?Log(1+(r*s)/b)/Log(r+1)
22.7940616057939

If you wanted, you could put this in A1, and copy down. However, the
equation doesn't look similar to the original.
=447000 - 147000*1.05^ROW()

This comes from the equation for the value at any given period. Period 22
would be:
n=22
?(b - b*(1 + r)^n + r*s)/r
16986.6741715063

n=23
?(b - b*(1 + r)^n + r*s)/r
-4513.9921199184

--
Dana DeLouis
Win XP & Office 2003


"sirsoto" <[email protected]> wrote in message
news:[email protected]...
>I want to make the following computation in the example cells below;
> Highlight the three cells where I entered my formula; Drag down until
> 300,000 equals zero (or near zero).
>
> a1=300,000-(49*150*1.05^0)
> a2=a2-(49*150*1.05^1)
> a3=a2-(49*150*1.05^2)
>
> The problem i am having is showing a series of exponential increases. I
> need 1.05 (which is 5%) to increase each year. Please advise. Sirsoto

I want to make the following computation in the example cells below;
Highlight the three cells where I entered my formula; Drag down until
300,000 equals zero (or near zero).

a1=300,000-(49*150*1.05^0)
a2=a2-(49*150*1.05^1)
a3=a2-(49*150*1.05^2)

The problem i am having is showing a series of exponential increases. I
need 1.05 (which is 5%) to increase each year. Please advise. Sirsoto

>>a2=a2-(49*150*1.05^1)<<

I'm assuming this is a typo and that cell A2 contains:

=A1-(49*150*1.05^1)

Change this to:

=A1-(49*150*1.05^(ROW()-1))

and drag down.

--

Vasant





I assume that the refernce in cell A2 is to A1 and not to A2.
"sirsoto" <[email protected]> wrote in message
news:[email protected]...
>I want to make the following computation in the example cells below;
> Highlight the three cells where I entered my formula; Drag down until
> 300,000 equals zero (or near zero).
>
> a1=300,000-(49*150*1.05^0)
> a2=a2-(49*150*1.05^1)
> a3=a2-(49*150*1.05^2)
>
> The problem i am having is showing a series of exponential increases. I
> need 1.05 (which is 5%) to increase each year. Please advise. Sirsoto

> ... Drag down until
> 300,000 equals zero (or near zero).

Hi. not sure if this would be helpful, but the number of periods to get to
zero could be given as an equation. For this, there is no integer solution.

s=300000 'starting amount
b=49*150
r=.05

?Log(1+(r*s)/b)/Log(r+1)
22.7940616057939

If you wanted, you could put this in A1, and copy down. However, the
equation doesn't look similar to the original.
=447000 - 147000*1.05^ROW()

This comes from the equation for the value at any given period. Period 22
would be:
n=22
?(b - b*(1 + r)^n + r*s)/r
16986.6741715063

n=23
?(b - b*(1 + r)^n + r*s)/r
-4513.9921199184

--
Dana DeLouis
Win XP & Office 2003


"sirsoto" <[email protected]> wrote in message
news:[email protected]...
>I want to make the following computation in the example cells below;
> Highlight the three cells where I entered my formula; Drag down until
> 300,000 equals zero (or near zero).
>
> a1=300,000-(49*150*1.05^0)
> a2=a2-(49*150*1.05^1)
> a3=a2-(49*150*1.05^2)
>
> The problem i am having is showing a series of exponential increases. I
> need 1.05 (which is 5%) to increase each year. Please advise. Sirsoto