Binomial quotient limit

$\displaystyle{\large\displaystyle \sum^{n}_{k=0}\dbinom{n}{k} = (1+1)^n = 2^n}$ $\large\displaystyle{\sum^{n/m}_{k=0}\dbinom{n}{mk} = \dfrac{(1+\phi)^n+(1+\phi^2)^n+\cdots +(1+\phi^{m-1})^n+(1+1)^n}{m}}$ $= \large\dfrac{(1+\phi)^n+(1+\phi^2)^n+\cdots +(1+\phi^{m-1})^n+2^n}{m}$ $\implies \dfrac{\displaystyle{\large\displaystyle \sum^{n}_{k=0}\dbinom{n}{k}}}{\large\displaystyle \sum^{n/m}_{k=0}\dbinom{n}{mk}} = \large\dfrac{m \times 2^n}{(1+\phi)^n+(1+\phi^2)^n+\cdots +(1+\phi^{m-1})^n+2^n}$ $\therefore \lim_{n \to \infty}{\dfrac{\displaystyle{\large\displaystyle \sum^{n}_{k=0}\dbinom{n}{k}}}{\large\displaystyle \sum^{n/m}_{k=0}\dbinom{n}{mk}} = \large\lim_{n \to \infty}{\dfrac{\left(\dfrac{m \times 2^n}{2^n}\right)}{\left(\dfrac{(1+\phi)^n+(1+\phi^2)^n+\cdots +(1+\phi^{m-1})^n+2^n}{2^n}\right)} = \boxed{m}}}$

Note: Here, $\large\phi$ is an $m^{th}$ root of unity.