The Compton effect problem Show that when a photon of energy E is scattered from a free electron at rest, the max kinetic energy of the recoiling electron is given by 
$$K_\text{max}=\frac{E^{2}}{E+\frac{mc^{2}}{2}}$$
Using $$\Delta \lambda = \frac{h}{mc}\left ( 1-\cos\phi \right )$$, the largest change in energy transfer from the photo to the electron, by conservation, would result in a larger change in wavelength.
Change in wavelength largest when $$\phi=\pi$$
Finally, the expression becomes 
$$\Delta \lambda=\lambda_{f}-\lambda_{i}=\frac{2h}{m_{e}c}$$
Stuck here.
Can someone give me a little help?
Edit:Tried substituting delta lambda into $$\Delta E=\frac{hc}{\Delta \lambda}$$ to figure out the change in energy which would be the energy gained by the electron after the collision but arrived at nothing similar to the required expression.
Edit:
I see what's the problem-technical algebraic error.
happy for this topic to be deleted .
 A: First of all, as this is a homework type quertion, I don't post full answer.  
You can't write $\Delta{E}=\frac{hc}{\Delta\lambda}$ because $\Delta\left( \frac{1}{\lambda}\right)=\frac{-1}{\lambda^2}\Delta\lambda$  
Now, you can simply write, from compton's equation,  $$\lambda'-\lambda=\frac{2h}{mc}\\
\implies \frac{1}{E'}-\frac{1}{E}=\frac{2}{mc^2}$$
This get when divide throuht by $hc$. Here, $E'$ and $\lambda'$ are energy and wavelength of defracted wave length. You can use conservation of energy to replace $E'$ in terms of $E$ and kinetic energy of electron. Put that into above equation and rearranging, you get the required relation.  
[edit]: as per comments, I add a little more.
You cannot write $\Delta{E}=\frac{hc}{\Delta\lambda}$ because , $E=\frac{hc}{\lambda}$ and so $\Delta{E}=\Delta\left(\frac{hc}{\lambda}\right)=\frac{-hc}{\lambda^2}\Delta\lambda$ .
And also the expression $\Delta{E}=\frac{hc}{\Delta\lambda}$ does not make any physical sense. Because this means that larger the difference between wavelength of incident and scattered photon, smaller the energy gained by the electron. And also, as the wavelength increases, energy decreases. So, there must be a negative sign.  
Hope you understand it.
