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# summations watch

1. how do you go about solving a summation of the form
(0.5)^(2n) from n=0 to infinity?
2. (Original post by pippabethan)
how do you go about solving a summation of the form
(0.5)^(2n) from n=0 to infinity?
Notice that by a law of indices, (0.5)^(2n) = (0.5^2)^n = 0.25^n. Now we can write out the summation as 1 + 0.25 + 0.25^2 + 0.25^3 + ..., which is a geometric series with first term 1 and common ratio 0.25. Thus its value, using the standard formula, is 1/(1-0.25) = 1/0.75 = 4/3.
3. (Original post by pippabethan)
how do you go about solving a summation of the form
(0.5)^(2n) from n=0 to infinity?
You need to use:

where a_n are the terms in the sum and r is the common ratio.

It might help to rewrite (0.5)^(2n) as (0.25)^n.
4. Thank you

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