HELP WITH MATHMATICS.

author=McBick
1 / .01%
2 / .03%
3 / .07%
4 / .15%
5 / .39%

This is almost : 1/.01%
2/.01 +(2^1)/100
3/.01+(2^1+2^2)/100
4/.01+(2^1+2^2+2^3)/100
but to keep it in this rather simple formula, you would have to have 31 and not 39, for 5, which would then be
5/.01 + (2^1+2^2+2^3+2^4)/100

or to have a regular fonction : if "n" is any number::
n/=n-1/ + ( 2^(n-1))/100

or again :
n/=1/ + {2^1 + 2^2+ +2^3+ .................. +2^(n-1)}/100
Does that help?

you're right, no its all divided by 100, not 10, sorry, my bad,but where you got it wrong is this :
Its :. 01 + (2^1)/100 = .01+2/100 =. 01+.02= .03


if you're looking for 2/ ( then 2 is n), and you only add to 1/ the exponentiations of 2 till n-1, the whole divided by 100, here : 2-1=1(n-1), so you'll only add 2^1/100. It even works for 0 : 1/= 0 +( 2^0)/100 (knowing anything ^0=1), so : 1/=0 + 1/100= 0 +.01=.01.

I rectified upper post.Does it help now ?

No, first there is no n^1, n is part of the exponent of 2, and its anything^0(not 1)=1
2^1=2, but 2^0=1.(I guess its sort of a mathematical convention, but/because proved right in all calculations.)

ok, good!