Question:

Find the sum of the infinite geometric series.?

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3-1+1/3-1/9+...

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  1. since it is subtracted then group the subtracted terms from the addition

    and we see the common ratio is 1/9 and starting term is 2

    ∏ 2* (1/9)^n

    use the infinite geometric series sum formula

    S(infinite) = a/1-r = 2/(1-1/9) = 2.25

    hope this helps


  2. Sum to infinity = a/(1-r), where a = first term and r = ratio

    In this case, a = 3 and r = -1/3

    So the sum to infinity = 3/(1 - (-1/3)) = 3/(4/3) = 9/4 = 2.25

    Note: this only works when |r| < 1, otherwise there's no sum to infinity

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