# Need help

Prove that

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

Note by Abdelfatah Teamah
1 year, 1 month ago

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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.

- 1 year ago

Thank you

- 1 year ago