Given a Poisson distribution, $2f(0) + f(2) = 2f(1)$, what is the mean of the distribution?

Question

If for a Poisson distribution $2f(0) + f(2) = 2f(1)$, what is the mean of the distribution?

I know that for X ~ POI($\lambda$), then the pdf for the random variable X is

\begin{equation} \text{P(X = x)} = \frac{e^{-\lambda} \lambda^{x}}{x!} \end{equation}

and its mean is

\begin{equation} \mu_x = \lambda. \end{equation}

Answer 1

$f(0)=e^{-\lambda}$, $f(1)=e^{-\lambda} \lambda$ , $f(2)=\frac{e^{-\lambda} \lambda^{2}}{2}$. Hence $2e^{-\lambda}+\frac{e^{-\lambda} \lambda^{2}}{2}=2e^{-\lambda} \lambda$, dividing by $e^{-\lambda}$, we get $\frac{\lambda^{2}}{2}-2\lambda+2=0$, hence $\lambda^{2}-4\lambda+4=0$, so $(\lambda-2)^{2}=0$ so $\lambda=2$

Answer 2

$$2f(0)+f(2)=2f(1)$$ $$e^{-\lambda}(2\lambda^0/0!+\lambda^2/2!)=e^{-\lambda}(2\lambda/1!)$$ $$2+\lambda^2/2=2\lambda$$

Solve the quadradic equation for $\lambda$ and you are done.