\(Z(^nC_p) = 1+ ^nC_1 z^{-1}+ ^nC_2 z^{-2} + ^nC_3 z^{-3} \)+……… which can be further equated to (1+z-1)n.

Explanation: δ(n) exists only at n=0 and δ(n+1) exists only at n=1. Therefore while substituting this function, the Z – Transform at every other place becomes zero except at n=1. Therefore the Z-Transform of δ(n+1) is \(\frac{1}{z}. \)

Explanation: Using the definition of Z- Transform we have \(∑_{n=0}^∞ (\frac{1}{n!} z^{-n})\). Now, expanding this we get \( 1+\frac{z^{-1}}{1} + \frac{z^{-2}}{2} + \frac{z^{-3}}{3} \)+ …………. This is nothing but the expansion of \(e^{\frac{1}{z}}\), hence the answer is \(\frac{1}{n!}.\)

Explanation: The inverse Z-Transform of \(\frac{z}{z-a} \) is \(a^n\). The inverse Z-Transform of \((\frac{z}{z-a})^2\) is the convolution of an and an. Now, the inverse Z-Transform of \((\frac{z}{z-a})^3\) is the convolution of the result of the previous step with an an. Thus we get the answer \(\frac{1}{2} . (n+1) (n+2) a^n U(n) \)