What is the sum of:
2 1 + 8 1 + 3 2 1 + 1 2 8 1 + . . .
if you added an infinite number of terms.
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S 4 1 S 4 3 S ⟹ S = 2 1 + 8 1 + 3 2 1 + 1 2 8 1 + ⋯ = 8 1 + 3 2 1 + 1 2 8 1 + 5 1 2 1 + ⋯ = 2 1 = 3 2
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If we notice clearly, this is a geometric sequence with the common ratio as 4 1 = r . Therefore since r is between − 1 and 1 , the sum to infinity exists. Using the formula of sum to infinity:
1 − r a where ’a’ is the first term
So we can substitute the values in our equation:
= 1 − 0 . 2 5 0 . 5 = 0 . 7 5 0 . 5 = 3 2