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Find the value of $\sum_{k=1}^{3n}\binom{6n}{2k-1}(-3)^k$.

My working:

\begin{align} & \sum\binom{6n}{2k-1}(-3)^k\\ =& \sum\binom{6n}{2k-1}(i\sqrt3)^{2k}\\ =& i\sqrt3\sum\binom{6n}{2k-1}(i\sqrt3)^{2k-1}\\ =& \frac{i\sqrt3[(1+i\sqrt3)^{6n}-(1-i\sqrt3)^{6n}]}{2}\\ =& \frac{i\sqrt3[(1+i\sqrt3+1-i\sqrt3)^{3n}((1+i\sqrt3-1+i\sqrt3)^{3n}]}{2}\\ =& \frac{i\sqrt3\cdot2^{3n}2^{3n}(i\sqrt3)^{3n}}{2} \\ =& \frac{2^{6n}(i\sqrt3)^{3n+1}}{2} \end{align}

But the correct answer is $0$ and I can't figure out how one gets that. Moreover, I want to know what's wrong with my solution.

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    show that your sum has the value zero – 2017-02-21
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    @Dr.SonnhardGraubner isn't that the question? How do I do that? – 2017-02-21
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    Line 5 is incorrect: $(a\pm b)^k\ne a^k\pm b^k$ in general and here as well. – 2017-12-27

2 Answers 2

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Hint: $$(1+i\sqrt 3)^{6n}-(1-i\sqrt 3)^{6n}=(re^{i\pi/3})^{6n}-(re^{-i\pi/3})^{6n}=r^{6n}(e^{i2\pi n}-e^{-i2\pi n})=r^{6n}(1-1)=r^{6n}\cdot 0$$

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    Okay I haven't studied complex numbers in such depth...can you elaborate? – 2017-02-21
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    @OsheenSachdev We knows, from Euler formula, that $$a+bi=r\cos(\alpha)+i\sin(\alpha)=re^{i\alpha}$$ from here, with the help of Pythagorean rule for rectangle triangles, we can know that $\alpha=\pi/3$ – 2017-02-21
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    $=re^{i\alpha}$, how? – 2017-02-21
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    Okay, thank you so much...but I was wondering why my solution doesn't simplify to zero...any idea? (its a product of nonzero numbers...) – 2017-02-21
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    @OsheenSachdev I dont understand what you did in your fourth step... we have that $$(1+i\sqrt 3)^2=1-3+2i\sqrt 3=-2+i\sqrt3\neq 2$$ – 2017-02-21
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    I just applied $a^2-b^2=(a+b)(a-b)$ – 2017-02-21
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    Oh right, that's wrong lol – 2017-02-21
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    @OsheenSachdev, I will add that we knows that $$r=|a+bi|=\sqrt{(a+bi)(a-bi)}=\sqrt{a^2+b^2}$$ In our case we have that $|1+i\sqrt3|=|1-i\sqrt3|=2=r$ (I deleted a previous comment accidentally). – 2017-02-21
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$\sum \binom{m}{2k-1}x^{2k-1} +\sum \binom{m}{2k}x^{2k} =\sum \binom{m}{k}x^k =(1+x)^m $.

$(1-x)^m =\sum \binom{m}{k}(-1)^kx^k $ so $(1+x)^m+(1-x)^m =\sum \binom{m}{k}x^k(1+(-1)^k) =2\sum \binom{m}{2k}x^{2k} $ and $(1+x)^m-(1-x)^m =\sum \binom{m}{k}x^k(1-(-1)^k) =2\sum \binom{m}{2k-1}x^{2k-1} $.

Therefore, putting $x = i\sqrt{3}$ and $m = 6n$,

$\begin{array}\\ \sum \binom{6n}{2k-1}(-3)^{k} &=\sum \binom{6n}{2k-1}(i\sqrt{3})^{2k}\\ &=i\sqrt{3}\sum \binom{6n}{2k-1}(i\sqrt{3})^{2k-1}\\ &=\frac{i\sqrt{3}}{2}((1-i\sqrt{3})^{6n}-(1+i\sqrt{3})^{6n})\\ &=0\\ \end{array} $

since $(1-i\sqrt{3})^3 =(1+i\sqrt{3})^3 =-8 $ and $(1-i\sqrt{3})^6 =(1+i\sqrt{3})^6 =64 $.

Similarly,

$\begin{array}\\ \sum \binom{6n}{2k}(-3)^{k} &=\sum \binom{6n}{2k}(i\sqrt{3})^{2k}\\ &=\frac{1}{2}((1-i\sqrt{3})^{6n}+(1+i\sqrt{3})^{6n})\\ &=\frac{1}{2}(64^n+64^n)\\ &=64^n\\ \end{array} $