Geometric Series

To find the sum S, of n terms of increasing powers:

S = 1 + r + r2 + . . . + rn-1

eliminate terms by evaluating the difference:

rS - S = (r + r2 + r3 + . . . + rn) - (1 + r + r2 + . . . + rn-1)

(r-1)S = rn - 1,  Therefore: S =   rn - 1   
   
 r - 1   

For an alternative solution, note that S is similar to the representation in base r of the number: 111...1  of n terms, each digit having the place value of a power of r.

If each digit of a number in base r had the value r-1, then adding 1 to this number would cause a carry to flow from the lowest digit to the highest, yielding a number with a 1 digit in the next higher place position followed by all 0 digits.  In a similar way:

(r-1)S = (r-1) + (r-1)r + (r-1)r2 + . . . + (r-1)n-1 + + + + + 1 = 1 --> r --> r2 --> . . . --> rn-1 --> rn

(r-1)S = rn - 1,  Therefore: S =   rn - 1   
   
 r - 1   

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