Show for all positive integers l : $0 = \sum_{k=0}^ {3l-1} (-3)^{k} { 6l \choose 2k+1}$

Question

Show for all positive integers l :

$$0 = \sum_{k=0}^ {3l-1} (-3)^{k} { 6l \choose 2k+1}$$

My teacher said that I should note "suitable Power of $ {\sqrt 3} + i$ " and use the binomial theorem to proof this.

Answer 1

Hint: Expand $(\sqrt3+i)^{6l}$ using the binomial theorem. (Where $6l$ is suggested by the top of that binomial coefficient in the problem statement.) Also, what is the value of $(\sqrt3+i)^{6}$? (Use the polar representation of $\sqrt3+i$.) And don't forget that two complex numbers are equal if and only if their real and imaginary parts both match.

Answer 2

We obtain for integral $l>0$: \begin{align*} \color{blue}{\sum_{k=0}^{3l-1}\binom{6l}{2k+1}(-3)^k} &=\frac{1}{i\sqrt{3}}\sum_{k=0}^{3l-1}\binom{6l}{2k+1}\left(i\sqrt{3}\right)^{2k+1}\tag{1}\\ &=\frac{1}{i\sqrt{3}}\sum_{k=0}^{6l}\binom{6l}{k}\left(i\sqrt{3}\right)^{k}\cdot\frac{1-(-1)^k}{2}\tag{2}\\ &=\frac{1}{2i\sqrt{3}}\left[\left(1+i\sqrt{3}\right)^{6l}-\left(1-i\sqrt{3}\right)^{6l}\right]\tag{3}\\ &=\frac{1}{2i\sqrt{3}}\left[(-8)^{2l}-(-8)^{2l}\right]\tag{4}\\ &\,\,\color{blue}{=0} \end{align*} and the claim follows.

Comment:

• In (1) we change the exponent of the term $(-3)^k$ from $k$ to $2k+1$ in order to adapt it with $\binom{6l}{\color{blue}{2k+1}}$.

• In (2) we add the even terms to the sum and multiply them with $0$ by $\frac{1-(-1)^k}{2}$ when $k$ is even. This way we prepare the sum for the binomial theorem in the next step. Note the upper limit is set to $6l$.

• In (3) we split the sum and apply the binomial theorem.

• In (4) we observe that $(1\pm i\sqrt{3})^3=-8$.