2.2+q14-18

These are all variations on the formulae

1/(1-z)=1+z+z^2+... and e^z=1+z/1!+z^2/2!+...

For example in q14: 1+z^2/1!+z^4/2!+... looks like the power series for e^z, but instead of z we have z^2. So the answer is e^(z^2). Similar idea in q15.

In q16 the series is the derivative of the series 1+(z-1)+(z-1)^2+...=1/(1-(z-1))

In q17 the series looks like e^z, except we have -(z-2pi i) instead of z. Hence the answer is e^(-z+2pi i)=e^(-z)

In q18, after you divide by z^2 you get the second derivative of 1+z+z^2+...=1/(1-z)