How do i evaluate 1+2+4+8+...+a subscript 10
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How do i evaluate 1+2+4+8+...+a subscript 10

[From: ] [author: ] [Date: 12-06-12] [Hit: ]
..a^0 + a^1 +a^2 +a^3 +..........
1 + 2 + 4 + 8 +......a subscript 10 is the same as
a^0 + a^1 +a^2 +a^3 +.....a^9 because there are 10 terms where a = 2.
This can be written as 2^0 + 2^1 + 2^2 +2^3 +......2^9
This is a geometric series similar to a^0 +a^1 +a^2 +a^3 +.....a^(n-1)
This is a geometric series whose sum of the first n terms is [a(1-r^n)]/(1-r) where r is the common ratio defined as the ratio between any term and its preceding term, and a is the first term
In our case, r = 2 and a = 2; Also n =10
Substituting we get the Sum = [2(1-2^(10)]/(1-2) = - 2*(1-1024) = 2* 1023 = 2046 which is the answer.
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