Sequence and series (11)

This is NOT my answer, I found it on Quora here and found it interesting and quite Brilliant

Let $\displaystyle S = \sum_{n=0}^\infty \binom{34}{4n}$

Using the property $\displaystyle \binom{n}{r} = \binom{n}{n-r}$ we have

$2S = \displaystyle \sum_{n=0}^\infty \left[\binom{34}{4n} + \binom{34}{4n-r}\right]$

Notice that this sum is $\begin{aligned}\left[{34\choose 0}+{34\choose 34}\right]+\left[{34\choose 4}+{34\choose 30}\right]+\left[{34\choose 8}+{34\choose 26}\right]+\ldots\tag*{}\end{aligned}$

Notice that as $n$ gets large, the terms are $0$ . With a little bit of rearranging: $\begin{aligned}2S=\sum_{n=0}^\infty\left[{34 \choose 4n}+{34 \choose 34-4n}\right]&=\sum_{n\text{ even}}^\infty{34\choose n}\tag*{}\end{aligned}$

Now, using Pascal's identity, ${n\choose r} = {n-1 \choose r-1} + {n-1\choose r}$ , we have that

$\begin{aligned}2S=\sum_{n\text{ even}}^\infty{34\choose n}&=\sum_{r=0}^\infty{33\choose r}=2^{33}\tag*{}\end{aligned}$ Thus, $\begin{aligned} S = \sum_{n=0}^\infty{34\choose 4n}=2^{32}\tag*{}\end{aligned}$

First note that $\dbinom nm = 0$ for $m>n$ . Therefore $\displaystyle \sum_{n=0}^\infty \binom {34}n = \sum_{n=0}^8 \binom {34}n$ By binomial theorem , we have:

$\small \begin{aligned} (1+x)^{34} & = \binom {34}0+ \binom {34}1 x + \binom {34}2 x^2 + \binom {34}3 x^3 + \binom {34}4 x^4 + \cdots + \binom {34}{32} x^{32} + \binom {34}{33} x^{33} + \binom {34}{34} x^{34} \\ (1+1)^{34} & = \binom {34}0 + \binom {34}1 + \binom {34}2 + \binom {34}3 + \binom {34}4 + \cdots + \binom {34}{32} + \binom {34}{33} + \binom {34}{34} \\ (1+i)^{34} & = \binom {34}0 + \binom {34}1 i- \binom {34}2 - \binom {34}3i + \binom {34}4 + \cdots + \binom {34}{32} + \binom {34}{33}i - \binom {34}{34} \\ (1-1)^{34} & = \binom {34}0 - \binom {34}1 + \binom {34}2 - \binom {34}3 + \binom {34}4 - \cdots + \binom {34}{32} - \binom {34}{33} + \binom {34}{34} \\ (1-i)^{34} & = \binom {34}0 - \binom {34}1 i - \binom {34}2 + \binom {34}3i + \binom {34}4 + \cdots + \binom {34}{32} - \binom {34}{33}i - \binom {34}{34} \end{aligned}$

$\small \begin{aligned} \implies (1+1)^{34} +(1+i)^{34}+(1-1)^{34}+(1-i)^{34} & = 4\left(\binom {34}0 + \binom {34}4 + \binom {34}8 + \cdots + \binom {34}{32} \right) \end{aligned}$

$\begin{aligned} \implies \sum_{n=0}^\infty & = \frac {(1+1)^{34} +(1+i)^{34}+(1-1)^{34}+(1-i)^{34}}4 \\ & = \frac {2^{34} +(2i)^{17}+0+(-2i)^{17}}4 \\ & = 2^{32} \end{aligned}$

Therefore, $k=\boxed {32}$ .