1

I'm stuck proving this result...

Let $a$ be a positive even integer and $m$ a positive integer.

Show that $$a^m\sum_{k=0}^{m}\frac{(-1)^k {2k \choose k}a^{-k}}{2^{2k}} $$ can be rewritten as $$\frac{2K+1}{2^b}$$ where $K\in \mathbb Z$ and $b$ satisifies $$b=m+e_b$$ where $e_b$ is the exponent of $2$ in $m!$

I really don't know how to begin... Thanks for your help.

Gabriel Romon
  • 35,428
  • 5
  • 65
  • 157

1 Answers1

1

We have $$a^m\sum^m_{k=0}\frac{(-1)^k\binom{2k}{k}a^{-k}}{4^k}=\frac{x}{2^{m+e}}$$

where $x$ is odd $$\sum^m_{k=0}\binom{2k}{k}(-4)^{-k}a^{m-k}=\frac{x}{2^{m+e}}=\frac{x2^{m-e}}{4^m}$$ If $m$ is even, then the parity of $-k$ is the same as $m-k$ and we can safely manipulate $$\sum^m_{k=0}\binom{2k}{k}(-4a)^{m-k}=x2^{m-e}\tag{1}$$ If not, the sign of the whole sum gets inverted, but this does not change the parity of it.By the same token as robjohn's answer, we have that $$\text{number of factors 2 in n!}=\nu_2(n) =\sum_{i=0}^kn_i\left(1+2+2^2+\dots+2^{i-1}\right)$$ Where $n_i$ are the i th coefficients of the binary representation of $n$. Since the binary representation of $2n$ is just the one of $n$ shifted one number to the left, we then have that $$ \begin{align} \nu_2(2n) &=\sum_{i=0}^kn_i\left(1+2+2^2+\dots+2^{i-1}+2^i\right)\\ &=\nu_2(n)+\sum_{i=0}^kn_i2^i=\nu_2(n)+n \end{align}$$ Then, the number of factors $2$ in the last term of $(1)$ is $$\nu_2(m)+m-2\nu_2(m)=m-v_2(m)=m-e$$

So now we just have to show that the other terms are an even multiple of $2^{m-e}$.

We then have that the exponent of two of the penultimate term is $$[m-1-\nu_2(m-1)]+2=m+1-\nu_2(m-1)$$ Since $\nu_2(m-1)\le\nu_2(m)$, we have $$m-\nu_2(m)\le m-\nu_2(m-1)<m+1-\nu_2(m-1)$$ Therefore the exponent of two in the last term is strictly smaller than the previous one,i.e. all the other terms are an even multiple of it. Finally, we know that $2^{m-e}$ is the highest power of $2$ that divides our expression.

Community
  • 1
chubakueno
  • 5,623