How to prove that $2^{2n+1}=\sum_{k=0}^{2n}(-1)^{k+n}\binom{4n+2}{2k+1}\;\;\;\forall n \in\mathbb{N}$

Question

I'm working on a related problem, which is to determine the Taylor series of $\sin(\pi x)\sinh(\pi x)$ evaluated at $x=0$.

I have succeeded in finding the series, but my coefficients are complicated sums: $\left (S_n=\sum_{k=0}^{2n}\frac{(-1)^{k+n}}{(2k+1)!(4n+1-2k)!}\right )$.

From my investigation, I have found that my formula generates the following sequence:

$S_1=1,\;S_2=\frac{1}{90},\;S_3=\frac{1}{113400},\;S_4=\frac{1}{681080400},\;S_5=\frac{1}{12504636144000}$

Plugging these denominators into the OEIS, we find the sequence A052277, which is directly related to my question: $\left (\sin(x)\times\sinh(x) =\sum_{n>=0}\;(-1)^n\times \frac{x^{4n+2}}{a(n)}\right )$ and it even states a fact I have independently proven: $\left ( S_n=\frac{\pi^{4n}}{\zeta(4n)} \right )$ which indirectly confirms that $S_n$ is the reciprocal of the terms in A052277.

This motivates me to multiply both sides by $(4n+2)!$ , leading to the identity I'm trying to prove:

$2^{2n+1}=\sum_{k=0}^{2n}(-1)^{k+n}\binom{4n+2}{2k+1}$

At this point, we can recall that the sum of all the coefficents in the n-th row of Pascal’s triangle equals $2^n$.So the intuition is that this alternating sum of thses coefficients in the $(4n+2)$-th row equals the full sum of coefficients in the $(2n+1)$th row.

With this intuition, I expressed $\binom{4n+2}{2k+1}$ as a sum of entries from the $(2n+1)$-th row:

$\binom{4n+2}{2k+1}=\sum_{j=0}^{2n+1}\binom{2n+1}{j}\binom{2n+1}{2k+1-j}$

However, I was unsuccessful in simplifying the full expression like so:

$\sum_{k=0}^{2n}(-1)^{k+n}\sum_{j=0}^{2n+1}\binom{2n+1}{j}\binom{2n+1}{2k+1-j}=\sum_{k=0}^{2n+1}\binom{2n+1}{k}$

Attempting to exchange the order of summation seemed promising, but did not lead to a simplification — or at least I haven't managed to find one.

I am therefore convinced that the formula in my question is true but remain unable to prove it. Can anyone suggest a method or identity that might help complete the proof?

Answer 1

I don't see any other solution except complex numbers

$$S_n=\sum_{k=0}^{2n}(-1)^{k+n}\binom{4n+2}{2k+1}=(-1)^n\sum_{k=0}^{2n}(-1)^k\binom{4n+2}{2k+1}$$

Identity from Newton's binomial theorem

$$(1+i)^{4n+2}=\sum_{j=0}^{4n+2}\binom{4n+2}{j}i^j=\sum_{k=0}^{2n}\binom{4n+2}{2k}i^{2k}+\sum_{k=0}^{2n}\binom{4n+2}{2k+1}i^{2k+1}$$

Since $i^{2k}=(-1)^k$, $i^{2k+1}=i(-1)^k$, we obtain a decomposition into real and imaginary parts

\begin{multline*} (1+i)^{4n+2}=\underbrace{\sum_{k=0}^{2n}\binom{4n+2}{2k}(-1)^k}_{\Re}+i\,\underbrace{\sum_{k=0}^{2n}\binom{4n+2}{2k+1}(-1)^k}_{\Im}\Rightarrow\\\Rightarrow \Im\big((1+i)^{4n+2}\big)=\sum_{k=0}^{2n}(-1)^k\binom{4n+2}{2k+1} \end{multline*}

Since $1+i=\sqrt{2}\,e^{i\pi/4}$, then

$$(1+i)^{4n+2}=(\sqrt{2})^{4n+2}\,e^{i(4n+2)\pi/4}=2^{2n+1}\,e^{i(n\pi+\pi/2)}=2^{2n+1}\,(-1)^n\,i$$

Therefore, $\Im\left( (1+i)^{4n+2} \right)=2^{2n+1}(-1)^n$, which means

$$S_n=(-1)^n\cdot\Im\big((1+i)^{4n+2}\big)=(-1)^n\cdot 2^{2n+1}(-1)^n=2^{2n+1}$$

Answer 2

There's a general approach to solving $$f(x)=\sum_{m=0}^Na_m\binom Nm x^m$$ when $a_m$ is periodic.

Here, after removing the common multile $(-1)^n,$ $a_m=0$ if $m$ even and $a_{m}=(-1)^{k}$ if $m=2k+1$ is odd. This has period $4.$

Here, it is periodic of period four, so the trick is to find $b_0,b_1,b_2,b_3$ such that:

$$b_0+b_1+b_2+b_3=a_0\\ b_0i^0+b_1i^1+b_2i^2+b_3i^3=a_1\\ b_0i^0+b_1i^2+b_2i^4+b_3i^6=a_2\\ b_0i^0+b_1i^3+b_2i^6+b_3i^9=a_3$$

Then $$f(x)=b_0(1+x)^{N}+b_1(1+ix)^N+b_2(1-x)^N +b_3(1-ix)^N.$$

Solving gives: $b_0=b_2=0,b_1=\frac1{2i},b_2=-\frac1{2i}.$

You get, setting $x=1,$

$$\sum_{m=0}^{N}a_m\binom Nm=\frac{(1+i)^N-(1-i)^N}{2i}$$

which is just the imaginary part of $(1+i)^N.$

With $N=4n+2,$ this is $(2i)^{2n+1}=2^{2n+1}(-1)^ni.$ But remember, we removed the $(-1)^n,$ so when we get that, the result is $2^{2n+1}.$

Answer 3

Notice that Alexander Shin's approach also applies to the original problem:

$$ \sin(\pi x)\sinh(\pi x) = i\sin(\pi(x))\sin(\pi(-ix)) = \frac{i}{2}\left(\cos(\pi(1+i)x)-\cos(\pi(1-i)x)\right)$$

is just the imaginary part of $-\cos((1+i)\pi x)$. Our function is an entire even function, so $[x^{2k+1}]\sin(\pi x)\sinh(\pi x)=0$, while

$$\begin{eqnarray*} [x^{2k}]\sin(\pi x)\sinh(\pi x)&=&-\operatorname{Im}[x^{2k}]\sum_{n\geq 0}\frac{(-1)^n x^{2n}\pi^{2n}(1+i)^{2n}}{(2n)!}\\&=&(-1)^{k+1}\frac{\pi^{2k}}{(2k)!}\operatorname{Im}\left((1+i)^{2k}\right)\\&=&(-1)^{k+1}\frac{\pi^{2k}}{(2k)!}\operatorname{Im}\left((2i)^{k}\right)\\&=&(-1)^{k+1}\frac{\pi^{2k}2^k}{(2k)!}\chi_4(k)\end{eqnarray*} $$

where $\chi_4(k)=(-1)^{\frac{k-1}{2}}$ if $k$ is odd and $\chi_4(k)=0$ if $k$ is even. This leads to

$$\sin(\pi x)\sinh(\pi x)=\sum_{j\geq 0}\frac{(-1)^j\, (2\pi^2)^{2j+1}}{(4j+2)!}\,x^{4j+2}.$$

Answer 4

Applying $$\sum_k a_{2k} = \sum_k \frac{1+(-1)^k}{2} a_k$$ with $a_k = i^k \binom{4n+2}{k+1}$ yields \begin{align} \sum_{k \ge 0} (-1)^{k+n} \binom{4n+2}{2k+1} &= (-1)^n \sum_{k \ge 0} i^{2k} \binom{4n+2}{2k+1} \\ &= (-1)^n \sum_{k \ge 0} \frac{1+(-1)^k}{2} i^k \binom{4n+2}{k+1} \\ &= \frac{(-1)^n}{2} \left(\frac{1}{i}\sum_{k \ge 0} i^{k+1} \binom{4n+2}{k+1} + \frac{1}{-i}\sum_{k \ge 0} (-i)^{k+1} \binom{4n+2}{k+1}\right) \\ &= \frac{(-1)^n}{2} \left(-i\left((1+i)^{4n+2}-i^{0} \binom{4n+2}{0}\right) + i\left((1-i)^{4n+2}-(-i)^{0} \binom{4n+2}{0}\right)\right) \\ &= \frac{(-1)^n}{2} \left(-i (1+i)^{4n+2} + i(1-i)^{4n+2}\right) \\ &= \frac{(-1)^n}{2} \left(2(-4)^n + 2(-4)^n\right) \\ &= 2^{2n+1} \end{align}