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Prove that

$\large 1 - ^{4n} C_1 - ^{4n} C_2 + ^{4n}C_3 + ^{4n}C_4 - ^{4n}C_5 - ^{4n}C_6 + \cdots + ^{4n}C_{4n} = (-4)^n .$

#Combinatorics

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Comments

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,

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

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.

Prasun Biswas - 4 years ago

Thank you

Abdelfatah Teamah - 4 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$