protons and electron are placed at point x=1cm, 3cm 5cm & x=2cm,4cm,6cm respectivelyon x-axis up to infinity what is potential difference origin

We know the potential from each charge is of the form

V=\frac{k\cdot Q}{r}

Assuming charges of +q for the protons and -q for the electrons, you'll get contributions from the protons as follows:

V_{protons}=\frac{k\cdot q}{1}+\frac{k\cdot q}{3}+\frac{k\cdot q}{5}+...

and contributions from the electrons as follows:

V_{electrons}=-\frac{k\cdot q}{2}-\frac{k\cdot q}{4}+\frac{k\cdot q}{6}+...

Summing all the contributions together, we get

V=\frac{k\cdot q}{1}-\frac{k\cdot q}{2}+\frac{k\cdot q}{3}-\frac{k\cdot q}{4}+... = k\cdot q\sum _{n=1}^{\infty}\frac{\left (-1 \right )^{n+1}}{n}=k\cdot q\cdot \ln \left (2 \right )

That last jump where the summation was replaced by ln(2) comes from the known value of the alternating harmonic series. See this link (http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)) for a more in-depth explanation.

By the way, since the charge locations are specified in cm, make sure your factor k is scaled to those units, not meters.