Why is the Rydberg formula giving me incorrect wavelengths? I am tasked with the following question:

Find the wavelengths of the first four members of the Brackett series in $\mathrm{He}^{+}$ – the set of spectral lines corresponding to transitions from $n \gt 4$ to $n = 4$.

In order to find the wavelengths it was my understanding that the relevant formula to use is
$$\frac{1}{\lambda}=R_{\infty}\left(\frac{1}{4^2}-\frac{1}{n^2}\right)\tag{1}$$
Rearranging,
$$\lambda=\frac{1}{R_{\infty}}\left(\frac{16n^2}{n^2-16}\right)$$
and taking the Rydberg constant $R_\infty$ to be $1.0974\times 10^7\text{m}^{-1}$
Then $n=8\rightarrow 4\implies \lambda\approx 1940\,\text{nm}$
and $n=7\rightarrow 4\implies \lambda\approx 2165\,\text{nm}$
and $n=6\rightarrow 4\implies \lambda\approx 2624\,\text{nm}$
and $n=5\rightarrow 4\implies \lambda\approx 4050\,\text{nm}$

The problem is that the answer is:

The wavelengths are given by $$\frac{1}{\lambda}=Rhc\left(\frac{1}{4^2}-\frac{1}{n^2}\right)$$ with $$Rhc=\frac{1}{91.1\,\text{nm}}$$ So the wavelengths are $1012\,\text{nm}$ ($n=5$ to $n=4$), $656\,\text{nm}$ ($n=6$ to $n=4$) and $541\,\text{nm}$
($n=7$ to $n=4$)


The formula that has been given in the answer I understood to be the change in energy $\Delta E$ when a transition takes place.
I am almost certain that $(1)$ is the correct formula to use (but it is apparently giving me the wrong results) unless there is something else I am missing.
Where am I going wrong?
 A: The general formula to calculate the wavelength of a hydrogen like species is $$\frac1\lambda=R_H\times z²\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)$$ where z is the atomic number of the species which in your case is equal to 2 for $He^+$
and $n_i$ and $n_f$ are the respective energy levels in the transition and $R_H$ is the Rydberg Constant for hydrogen atom.
A: The Rydberg formula has to be adjusted to that the nuclear charge $Z=2$. For hydrogen like ions (single electron + nucleus), the electron energies are proportional to $Z^2$, so you should multiply $R_{\infty}$ by four.
A: Okay so just a wild guess at why your book says $Rhc$ instead of $R$ as you mentioned in the comments.  It might be $R_Hc$ where $R_H$ is the Rydberg's constant and $c$ is the speed of light. 
I believe it could be because it's dealing with the frequency on the left hand side? Because we know that $\frac{c}{\lambda}$ = $\nu$ (where the symbols have their usual meanings).
Your book might want to say that 

$$\nu= R_H\cdot c\left(\frac{1}{{n_i}^2}−\frac{1}{{n_f}^2}\right)Z^2$$

The symbols for frequency and wavenumber are very similar. The first is $\nu$ and the other is $\bar\nu$
