Rewriting the Multi-Sum Equation: Simplifying f(x)=1+R(x)

[EngWiPy]

Hello,

I have the following equation:

$$f(x)=\sum_{r=m}^{M_B}\,\sum_{i=0}^{M_B-r}\,\sum_{j=0}^{r+i}\,\sum_{k=0}^{j(N_B-1)}(-1)^{i+j}{M_B\choose r}{M_B-r\choose i}{r+i\choose j}\left(\frac{x}{C}\right)^k\,e^{jx/C}$$

and I want to write it in the form of $$f(x)=1+R(x)$$

$$m$$ will be any number from 1 up to $$M_B$$, and $$f(x)=1-e^{x/C}$$ for $$M_B=N_B=1$$.

Can anyone help me, please? Because I am not sure about the limits when extracting some values of them.

Regards

 

[MathematicalPhysicist]

$$f(x)=1+\sum_{r=m}^{M_B-1}\,\sum_{i=1}^{M_B-r}\,\sum_{j=1}^{r+i}\,\sum_{k=1}^{j(N_B-1)}(-1)^{i+j}{M_B\choose r}{M_B-r\choose i}{r+i\choose j}\left(\frac{x}{C}\right)^k\,e^{jx/C}$$.

 

[EngWiPy]

$$f(x)=1+\sum_{r=m}^{M_B-1}\,\sum_{i=1}^{M_B-r}\,\sum_{j=1}^{r+i}\,\sum_{k=1}^{j(N_B-1)}(-1)^{i+j}{M_B\choose r}{M_B-r\choose i}{r+i\choose j}\left(\frac{x}{C}\right)^k\,e^{jx/C}$$.

But this doesn't give us $$f(x)=1-e^{x/C}$$ for $$M_B=N_B=1$$. And we have the problem for the last summation where the limits will be 1 to 0, how to handle this problem?

 

[EngWiPy]

Where are the mathematician? I really need this one. Please help if you can.