Is \( 2 + 2^2 + 2^3 + 2^4 + ... + 2^{10}\) divisible by \(3\)?

It seems undetermined, because there were no threes, but let's try to pair it, extract it to find the threes.

\(A = (2 + 2^2) + (2^3 + 2^4)+...+ (2^9 + 2^{10})\)

\(A = 2(1+2) + 2^3(1+2) + ... + 2^9(1+2)\)

\(A = 2 \times 3 + 2^3 \times 3 + ... + 2^9 \times 3\)

\(A = (2 +2^3 + ... + 2^9) \times 3\)

Look! The series has x3! So A is divisible by 3.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

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}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestYes it is divisible!!!!!!!!!!!!!

Log in to reply

An easier way to prove that it is divisible by \(3\) is by using modular arithmetic. Let the given expression be \(Z\).

\[Z=\sum_{i=1}^{10} 2^i \equiv \sum_{i=1}^{10} (-1)^i\pmod{3}\\ \implies Z\equiv \left(\sum_{i=1}^5 (-1)^{2i}\right)+\left(\sum_{i=1}^{5} (-1)^{2i-1}\right)\pmod{3}\\ \implies Z\equiv \left(\sum_{i=1}^5 (1)\right)+\left(\sum_{i=1}^5 (-1)\right)\equiv \sum_{i=1}^5 (1-1)\pmod{3}\\ \implies Z\equiv \sum_{i=1}^5 (0)\pmod{3}\\ \implies Z\equiv 0\pmod{3}\\ \implies 3\mid Z\]

where \(a\mid b\) denotes that \(a\) divides \(b\), or in other words, \(b\) is divisible by \(a\).

Log in to reply

I'm an 7th grade student, so I just use original numbers :)) Sorry, can't understand.

Log in to reply

It's nothing much complicated. You can review the modular arithmetic wikis or ask me which part of it you can't understand?! I'd be more than glad to help you out. :)

Log in to reply

Log in to reply

\[S=\sum_{i=1}^6 2^i\]

The initial and final values of \(i\) are given in lower and upper sides of the "sigma" symbol and the value of \(i\) gets incremented in each summation step.

Log in to reply