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Prove that

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

Note by Abdelfatah Teamah
8 months 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.

Prasun Biswas - 8 months ago

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Thank you

Abdelfatah Teamah - 7 months, 4 weeks ago

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