MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems-Fall 2003 PROBLEM SET 1 SOLUTIONS (E1)(O&W154) (a) For the r=l case, we have 1+1+12 n=0 For the r+ 1 case, by carrying out the long division, we can see that 1-r 1 N b) Using the formula we just derived for the r+ 1 case, we have 1 If r lim n=0� � � � MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems—Fall 2003 Problem Set 1 Solutions (E1) (O&W 1.54) (a) For the r = 1 case, we have: n=N−1 1 = 1 + 11 + 12 + + 1N−1 · · · n=0 = N For the r =→ 1 case, by carrying out the long division, we can see that 1 r = 1 + r + r1 + r2 + + rN−1 + N 1 − r · · · 1 − r N �−1 Nr n = r + 1 − r n=0 N �−1 1 − rN nr = 1 − r n=0 (b) Using the formula we just derived for the r =→ 1 case, we have � N−1 nr = lim rn N�� n=0 n=0 1 − rN = lim N�� 1 − r 1 rN = 1 − r − lim N�� 1 − r If | | r < 1 Nr lim = 0 N�� 1 − r So, � 1 nr = 1 − r n=0 1