How can I demonstrate that the decimal equivalent for a binary number consisting of $n$ $1$'s is $2^n-1$?

Question

I know that $2^n-1$ gives you the decimal equivalent for any word consisting of $n$ $1$'s, but how can I demonstrate that in general terms for any word of $n$ $1$'s?

It's nozt the decimal equivalent, it's simply the number – Hagen von Eitzen Sep 09 '13 at 20:04 — Hagen von Eitzen, Sep 09 '13 at 20:04

score 1 · Answer 1 · answered Sep 09 '13 at 20:03

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Sum the implicit geometric series:

$$\underbrace{111\ldots111}_n=\sum_{k=0}^{n-1}2^k=\;?$$

answered Sep 09 '13 at 20:03

Brian M. Scott

631,399

score 1 · Answer 2 · answered Sep 09 '13 at 20:08

1

Hint: what is the result if you add $1$ to any such word? What does that tell you?

answered Sep 09 '13 at 20:08

Ben Grossmann

234,171
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How can I demonstrate that the decimal equivalent for a binary number consisting of $n$ $1$'s is $2^n-1$?

2 Answers2