Binomial coefficients with step

Question

I know how to find $$\sum_{k=0}^{\lfloor n/2\rfloor}{{n}\choose{2k}} = 2^{k-1}.$$

But I have no idea how to find $\sum_{k=0}^{\lfloor n/3\rfloor}{{n}\choose{3k}}$ and $\sum_{k=0}^{\lfloor n/4\rfloor}{{n}\choose{4k}}$. Some hint could be useful.

Answer 1

Let $z_1=e^{2\pi i/3}$ and $z_2=e^{-2i\pi/3}$, then $$(1+1)^n + (1+z_1)^n + (1+z_2)^n=2^n +2\cos(n\pi/3)=3\sum_{k=0}^{\lfloor n/3\rfloor}{{n}\choose{3k}}$$ Hence $$\sum_{k=0}^{\lfloor n/3\rfloor}{{n}\choose{3k}}=\frac{2^n+2\cos(n\pi/3)}{3}= \frac{2^n+a_n}{3}$$ where $a_n$ is the 6-periodic sequence $\{2, 1, -1, -2, -1, 1\}$.

The details can be found here: Roots of unity filter, identity involving $\sum_{k \ge 0} \binom{n}{3k}$

By using a similar method, it turns out that for $n\geq 1$, $$\sum_{k=0}^{\lfloor n/4\rfloor}{{n}\choose{4k}}=2^{n-2}+2^{(n-2)/2}\cos(n\pi/4).$$ See the nice answer of robjohn.

Answer 2

If $\omega=e^{2\pi i/3}$, then $$ \frac{1+\omega^n+\omega^{-n}}3=\left\{\begin{array}{} 1&\text{if }n\equiv0\pmod3\\ 0&\text{if }n\not\equiv0\pmod3\\ \end{array}\right. $$ Therefore, $$ \begin{align} \sum_{k=0}^{\lfloor n/3\rfloor}\binom{n}{3k} &=\sum_{k=0}^n\binom{n}{k}\frac{1+\omega^k+\omega^{-k}}3\\[4pt] &=\frac{2^n}3+\frac{(1+\omega)^n}3+\frac{\left(1+\omega^{-1}\right)^n}3\\[4pt] &=\frac{2^n}3+\frac{e^{\pi in/3}}3+\frac{e^{-\pi in/3}}3\\[4pt] &=\frac{2^n+2\cos\left(\frac{\pi n}3\right)}3 \end{align} $$ Note that $2\cos\left(\frac{\pi n}3\right)=\{2,1,-1,-2,-1,1\}$ when $n\equiv\{0,1,2,3,4,5\}\pmod6$.

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$$ \frac{1+i^n+(-1)^n+(-i)^n}4=\left\{\begin{array}{} 1&\text{if }n\equiv0\pmod4\\ 0&\text{if }n\not\equiv0\pmod4\\ \end{array}\right. $$ Therefore, $$ \begin{align} \sum_{k=0}^{\lfloor n/4\rfloor}\binom{n}{4k} &=\sum_{k=0}^n\binom{n}{k}\frac{1+i^k+(-1)^k+(-i)^k}4\\[4pt] &=\frac{2^n}4+\frac{(1+i)^n}4+\frac{0^n}4+\frac{(1-i)^n}4\\[4pt] &=\frac{2^n}4+\frac{2^{n/2}e^{\pi in/4}}4+\frac{0^n}4+\frac{2^{n/2}e^{-\pi in/4}}4\\[4pt] &=\frac{2^n+[n=0]+2^{n/2}\cdot2\cos\left(\frac{\pi n}4\right)}4 \end{align} $$ Note that $2\cos\left(\frac{\pi n}4\right)=\left\{2,\sqrt2,0,-\sqrt2,-2,-\sqrt2,0,\sqrt2\right\}$ when $n\equiv\{0,1,2,3,4,5,6,7\}\pmod8$.

Answer 3

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I believe that @RobertZ and @robjohn answers provide the straightforward and simpler methods. However, it will be useful to see another point of view which is somehow related to the above mentioned answers.

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With $\ds{\ell = 1,2,3,\ldots}$ \begin{align} \color{#f00}{\sum_{k = 0}^{\lfloor\,{n/\ell}\,\rfloor}{n \choose \ell k}} &\ =\ \sum_{k = 0}^{\infty}\ \overbrace{ \oint_{\verts{z}\ =\ 1^{+}}\,\,{\pars{1 + z}^{n} \over z^{\ell k + 1}} \,{\dd z \over 2\pi\ic}}^{\ds{n \choose \ell k}}\ =\ \oint_{\verts{z}\ =\ 1^{+}}\,\,{\pars{1 + z}^{n} \over z} \sum_{k = 0}^{\infty}\pars{1 \over z^{\ell}}^{k}\,{\dd z \over 2\pi\ic} \\[5mm] & = \oint_{\verts{z}\ =\ 1^{+}}\,\,{\pars{1 + z}^{n} \over z} {1 \over 1 - 1/z^{\ell}}\,{\dd z \over 2\pi\ic} = \oint_{\verts{z}\ =\ 1^{+}}\,\, {z^{\ell - 1}\,\,\pars{1 + z}^{n} \over z^{\ell} - 1}\,{\dd z \over 2\pi\ic} \end{align}

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The integrand has $\ds{\ell}$ simple poles which are the roots $\ds{\braces{r}}$ of $\ds{z^{\ell} - 1 = 0}$. Then, \begin{align} \color{#f00}{\sum_{k = 0}^{\lfloor\,{n/\ell}\,\rfloor}{n \choose \ell k}} & = \sum_{r}{r^{\ell - 1}\,\,\pars{1 + r}^{n} \over \ell r^{\ell - 1}} = \color{#f00}{{1 \over \ell}\sum_{r}\pars{1 + r}^{n}}\,,\qquad \ell = 1,2,3,\ldots \end{align}

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For instance, $$ \left\lbrace\begin{array}{rclcl} \ds{\sum_{k = 0}^{n}{n \choose k}} & \ds{=} & \ds{{1 \over \color{#f00}{1}}\pars{1 + 1}^{n}} & \ds{=} & \ds{2^{n}} \\[2mm] \ds{\sum_{k = 0}^{\left\lfloor\,{n/\color{#f00}{2}}\,\right\rfloor} {n \choose \color{#f00}{2}k}} & \ds{=} & \ds{{1 \over \color{#f00}{2}}\bracks{\pars{1 + 1}^{n} + \pars{1 - 1}^{n}}} & \ds{=} & \ds{2^{n - 1}\,\, +\,\, \half\,\delta_{n0}} \\[2mm] \ds{\sum_{k = 0}^{\left\lfloor\,{n/\color{#f00}{3}}\,\right\rfloor} {n \choose \color{#f00}{3}k}} & \ds{=} & \ds{{1 \over \color{#f00}{3}}\sum_{k = -1}^{1}\pars{1 + \exp\pars{2\pi k\ic \over 3}}^{n}} & \ds{=} & \ds{{1 \over 3}\bracks{2^{n} + 2\cos\pars{n\,{\pi \over 3}}}} \\[5mm] \ds{\sum_{k = 0}^{\left\lfloor\,{n/\color{#f00}{4}}\,\right\rfloor} {n \choose \color{#f00}{4}k}} & \ds{=} & \ds{{1 \over \color{#f00}{4}}\bracks{\pars{1 + 1}^{n} + \pars{1 - 1}^{n} + \pars{1 + \ic}^{n} + \pars{1 - \ic}^{n}}} && \\ & \ds{=} & \ds{{1 \over 4}\bracks{2^{n} + \delta_{n0} + 2^{n/2 + 1}\,\cos\pars{n\,{\pi \over 4}}}}&& \end{array}\right. $$