$f\left(x\right)=2x^n+\lambda ,f\left(4\right)=133,f\left(5\right)=255$

$f\left(4\right)=2.4^{\mathrm{n}}+\lambda =133 \dots \dots \dots \left(\mathrm{i}\right)$

$f\left(5\right)=2.5^{\mathrm{n}}+\lambda =255 \dots \dots \dots \left(\mathrm{ii}\right)$

$\therefore f\left(5\right)-f\left(4\right)=2\left(5^n-4^n\right)=122$

$\Rightarrow 5^n-4^n=5^3-4^3\Rightarrow n=3$

$\therefore f\left(4\right)=133=2\times 4^3+\lambda$

$\therefore \lambda =5$

$\therefore f\left(x\right)=2x^3+5$

$\therefore f\left(3\right)=59\text{ and }f\left(2\right)=21$

$\therefore f\left(3\right)-f\left(2\right)=38=2\times 19=2^a{19}^b,(a=1,b=1)$

$\text{ Sum of factors }f\left(3\right)-f\left(2\right)=\frac{2^{a+1}-1}{2-1}·\frac{{19}^{b+1}-1}{19-1}=\frac{3\times 360}{18}=60$