Geometric Series
To find the sum S, of n terms of increasing powers:
S = 1 + r + r^{2} + . . . + r^{n-1}
eliminate terms by evaluating the difference:
rS - S = (r + r^{2} + r^{3} + . . . + r^{n}) - (1 + r + r^{2} + . . . + r^{n-1})
(r-1)S = r^{n} - 1,_{ } Therefore: S = | r^{n} - 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)r^{2} + . . . + (r-1)^{n-1} + + + + + 1 = 1 --> r --> r^{2} --> . . . --> r^{n-1} --> r^{n}
(r-1)S = r^{n} - 1,_{ } Therefore: S = | r^{n} - 1 | |
r - 1 |
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