+2a2+3a3 a(1+2a+3a Now we can seperate the contents of the paranthesis on the right-hand-side(rHs) of the equation above as follows a(1+a+a2+…+a+2a2+3a3+…) (1+a+ )+a(a+2a2+3a3+…) Note that the contents of the second paranthesis on the RhS is the very expression we are trying to evaluate (1-a)∑ n=0 Using the result from part(b)for a<1 a�� � �� � � � � �� �� � � �� � = � �� �� � �� � (c) n�n = � + 2�2 + 3�3 + · · · n=0 = � 1 + 2� + 3�2 + � · · · Now we can seperate the contents of the paranthesis on the right-hand-side (RHS) of the equation above as follows: n�n = � 1 + � + �2 + + � + 2�2 + 3�3 + � · · · · · · n=0 = � 1 + � + �2 + � + � � + 2�2 + 3�3 · · · + · · · Note that the contents of the second paranthesis on the RHS is the very expression we are trying to evaluate: n�n = � 1 + � + �2 + + n�n · · · n=0 n=0 (1 − �) n�n = � 1 + � + �2 + � · · · n=0 �n n=0 Using the result from part (b) for | | � < 1, (1 − �) n�n = 1 − � n=0 n�n = (1 − �)2 n=0 2