Matematické Fórum

stuart clark · 20. 01. 2020 11:05

The sum of series

$\bigg[\bigg(\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\cdots \bigg)-\frac{1}{2}\bigg(\binom{n}{1}+\binom{n}{2}+\binom{n}{4}+\binom{n}{5}+\cdots \bigg)\bigg]^2+\frac{3}{4}\bigg[\binom{n}{1}-\binom{n}{2}+\binom{n}{4}-\binom{n}{5}+\cdots \bigg]^2$

stuart clark · 23. 01. 2020 14:30

Thanks friends got it

Using $\displaystyle (1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\cdots +\binom{n}{n}x^n.$

Now put $\displaystyle x=\omega=-\frac{1}{2}+i\frac{\sqrt{3}}{2}.$

$\displaystyle (1+\omega)^n=\binom{n}{0}+\binom{n}{1}\cdot \omega+\binom{n}{2}\cdot \omega^2+\binom{n}{3}\cdot \omega^3+\binom{n}{4}\cdot \omega^4+\cdots$

$\displaystyle \bigg(\frac{1}{2}+i\frac{\sqrt{3}}{2}\bigg)^n=\bigg[\bigg(\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\cdots\bigg)-\frac{1}{2}\bigg(\binom{n}{1}+\binom{n}{2}+\binom{n}{4}+\cdots\bigg)\bigg]-i\frac{\sqrt{3}}{2}\bigg[\binom{n}{1}-\binom{n}{2}+\binom{n}{4}-\binom{n}{5}+\cdots\bigg]$

Now taking modulus on both side, We have

$\displaystyle \bigg[\bigg(\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\cdots \bigg)-\frac{1}{2}\bigg(\binom{n}{1}+\binom{n}{2}+\binom{n}{4}+\binom{n}{5}+\cdots \bigg)\bigg]^2+\frac{3}{4}\bigg[\binom{n}{1}-\binom{n}{2}+\binom{n}{4}-\binom{n}{5}+\cdots \bigg]^2=1.$

Matematické Fórum

#1 20. 01. 2020 11:05

Binomial sum

#2 23. 01. 2020 14:30

Re: Binomial sum

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