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Old 08-28-2002, 10:57 AM   #9 (permalink)
shahani
 
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Simple algebra really:

1+r+...r**n=r(1/(r+1)+r+...r**(n-1)) - Factor out r as Cadd says.

Divide by (1-r)/(1-r) ---This is effectively 1 so doesn't change anything.

[r(1/(r+1)+r+...r**(n-1))] x (1-r)/(1-r)=

{[r(1/(r+1)+r+...r**(n-1))]-[r+r**2.....+r**(n+1)]}/(1-r)

Above you have just "opened up the brackets and multiplied by 1 and by -r.


={(1+r+r**2....+r**n)-(r+r**2+ .....r**(n+1)}/(1-r)

Cancell all + and - terms above:


{1-r**(n+1)}/(1-r)

QED

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