Notice that since \(\dbinom{4n}r=\dbinom {4n}{4n-r}\), the terms \(\dbinom{4n}k\) with \(k\equiv 1\pmod 4\) and \(k\equiv -1\pmod 4\) cancel out each other. The resulting series is,

Now notice that if \(\omega\) be a primitive \(4^{\textrm{th}}\) root of unity, say \(\omega=i\), where \(i\) is the imaginary unit (square root of -1), then the above series equals \(\frac 12\left((1+\omega)^{4n}+(1-\omega)^{4n}\right)\).

With \(\omega=i\), this equals \(\frac 12\left((1+i)^{4n}+(1-i)^{4n}\right)\). Since \((1+i)^4=(1-i)^4=-4\), we have that the series equals \(\frac 12\left((-4)^n+(-4)^n\right)=(-4)^n\), and we're done.

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:

TopNewestNotice that since \(\dbinom{4n}r=\dbinom {4n}{4n-r}\), the terms \(\dbinom{4n}k\) with \(k\equiv 1\pmod 4\) and \(k\equiv -1\pmod 4\) cancel out each other. The resulting series is,

\[\binom{4n}0-\binom{4n}2+\dbinom{4n}4-\ldots-\binom{4n}{4n-2}+\binom{4n}{4n}=\sum_{k=0}^{2n}(-1)^k\binom{4n}{2k}\]

Now notice that if \(\omega\) be a primitive \(4^{\textrm{th}}\) root of unity, say \(\omega=i\), where \(i\) is the imaginary unit (square root of -1), then the above series equals \(\frac 12\left((1+\omega)^{4n}+(1-\omega)^{4n}\right)\).

With \(\omega=i\), this equals \(\frac 12\left((1+i)^{4n}+(1-i)^{4n}\right)\). Since \((1+i)^4=(1-i)^4=-4\), we have that the series equals \(\frac 12\left((-4)^n+(-4)^n\right)=(-4)^n\), and we're done.

Log in to reply

Thank you

Log in to reply