Resolved: "How do you prove that10n−110n−2=1+∑i=0∞2i×10−n(i+1)"

Bradlee Hooper

Answered question

2022-02-27

How do you prove that

\frac{10^{n} - 1}{10^{n} - 2} = 1 + \sum_{i = 0}^{\infty} 2^{i} \times 10^{- n (i + 1)}

Elijah Hunt

Beginner2022-02-28Added 5 answers

Your sum is a geometric series:

\sum_{i = 0}^{\infty} 2^{i} \times 10^{- n (i + 1)} = 10^{- n} \sum_{i = 0}^{\infty} {(\frac{2}{10^{n}})}^{i}

= 10^{- n} \times \frac{1}{1 - 2 \times 10^{- n}} = \frac{1}{10^{n} - 2}

You are done once you observe that

\frac{10^{n} - 1}{10^{n} - 2} = \frac{10^{n} - 2}{10^{n} - 2} + \frac{1}{10^{n} - 2}

= 1 + \frac{1}{10^{n} - 2}

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