I would rewrite this as a sequence, where \(a_{1}=1+2, a_{2}=1+2(1+2)\) and so on. finding the values of each term would give you 3,7,15,31... and this ends up being \(a_{n}=2^{n+1} -1\). Therefore, the value you are finding for this problem would be \(2^{2018}-1\). Hope this is correct and if it's not, tell me and I will revise it :D

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

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`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

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TopNewestThe generel form would be -( 2^(n+2) ) -1 where n is the number of brackets

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I would rewrite this as a sequence, where \(a_{1}=1+2, a_{2}=1+2(1+2)\) and so on. finding the values of each term would give you 3,7,15,31... and this ends up being \(a_{n}=2^{n+1} -1\). Therefore, the value you are finding for this problem would be \(2^{2018}-1\). Hope this is correct and if it's not, tell me and I will revise it :D

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Oh, now I get it. You are right :) :)

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Thanks and I am so glad that you got it!

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2^2018-1

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How can i find it? Please describe me

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