What does $1 + 2 + 4 + 8 + 16 + 32 + ...$ equal?

You could rewrite the equation as $(2 - 1)(1 + 2 + 4 + 8 + 16 + 32 + ...)$

This is then written as $(2 + 4 + 8 + 16 + 32 + ...) - 1 - 2 - 4 - 8 - 16 - 32 - ...$.

Which results in $1 + 2 + 4 + 8 + 16 + 32 + ...= -1$!!!.

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## Comments

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TopNewestThis is an infinite geometric series which has ratio $r=2>1$. So this doesn't converge, and as the result we get an non-sense answer.

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I'm sorry but you can't just use the distributive laws when dealing with infinitely many terms. A simple proof that this is wrong is that a sum diverges when its partial sums diverges. Clearly the partial sums are $2^n-1$ which approaches to infinity as $n$ does, therefore the sum does not exist and is undefined.

Or instead you are talking about something much more complicated like in a different number system, like 2-adic numbers. :P

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Damn, we communicated before Schol of Excellence???

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I do think this answer is a bit weird, but the method seems correct, since it is an

infiniteseries! and does not end.Log in to reply

Just an illusion.....,

Confine up to 5 or 6 terms and try to evaluate..

## InitiallyTrolled me too..,

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I had a similar derivation which I am not so sure about. My first note on Summations

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these laws are not valid when we are dealing with infinite sums.

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ahh I like these fallacious proofs - they really test your observation. I think the reason why you got -1 was because the factorization only works if the expression is finite.

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It has to do with the fact that you're ignoring the last term, which is infinitely large and therefore very important

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There is no largest term.

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I do not see where $-1$ came from.

EDIT: For a proof of where this summation went wrong, check my comment down here.

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Remove the brackets and you get 2+4+8+16+32+...-1-2-4-8-16-32. The +2 and -2 cancel each other out, so does +4 and -4, +8 and -8, etc. until all that's left is -1

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This is obviously false, we still have a term in the first sequence missing.

If the RHS

ends, let's say, in $2^n$, the LHS will end in $2^{n+1}$. This leads us that the $\Sigma = 2^{n+1} - 1$, which is true.But we cannot evaluate a limit for $n \rightarrow \infty$, because $\Sigma$ would tend also to infinity.

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$(2 - 1)(1 + 2 + 4 + 8 + 16 + 32 + ...)$, you just simplify it. $2(1 + 2 + 4 + 8 + ...) -1(1 + 2 + 4 + 8 + 16 +...)$ so it is $2 + 4 + 8 + 16 + 32 ...- 1 - 2 - 4 - 8 - 16...$. $2$ and $-2$, $4$ and $-4$, etc. all cancel each other out until $-1$ remains

Since it wasLog in to reply

$1+2+3+4+5+6+\ldots\infty=\frac{-1}{12}$, requires the assumption that the R.H.S.

However, the proof fordoes notend. The point of infinite sequences is that they do not end. Even the proof for $9.99999\ldots=10$ requires the $9$s to never end.Log in to reply

Nice logic!!

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