Question

Instructions: Please show all of your work. Unsupported answers may receive no credit. 1. (20 pts) Use mathematical induction to show that for integers n 21, 2.21 +3.22 + ... + (n + 1)21 = n. 21+1 w 2. (20 pts) Let P(n) be statement that a postage of n cents can be formed using only 4-cent and 7-cent stamps. Using strong induction, prove P(n) is true for n 2 18.

Answer 1

Now, Ewn (3) shows that result (1) IS TRUE Sar n=k+1. Q!1; Hore viven statement P (n) = 2.21 +3,22 +--+ (n+1.2n = n.ant, no wring mathematical Induction method, STEP-I!- Far n=1 L.4.5 = 2.21 = 4 and R. His = 1,2 = 4 L.H.S = R.H.S=u Thus Result is true far n=1. STEP-HI! - Let this result is true far n=k. then D(K) = 2,21 +3:22+...+ (x+11.2K = K. akti STEP-u! - Now, we have to show hill Result is truu far. n = 1 + 1. Now, By (2) we have 2.21 +3.22 adding (K+2). 2 K +1 on both sides, we get 2.21 +3.22 + (x+1). 2K+ (1+2), 2K +1 +(K+2)-2 (2x+2).2K+1 KAI (K+1). 2 K+2 2.21+3.2? + -. + (1+2).2 - (2) + ...t (K11).2k K. 2K+1 X+1 t - - Kipkt -13) Thuss, the result 2.21 +3, 2²+ .-- + (1+1)-2h = n. ant! is TRUE &no

9.2 Here; Pen) = a postage of n-cents can be formed wing only y-cent and 7- cent stamp. n. 7,18 Now, uring strong Induction method. STEP-I! - Far n =18 we have 2x7 + 4 =18 = P(18) is true Far n=19 we have 7 + 3x4 = 19 = Pera) is also true. STEP-Il- Arsume that P118), P119) .. PIK) are all true. STEP-H! we have to show that P (H+1) is also true. Now K+1= (K-3) tu -(1) Also (K-3) can be formed by u cent and 7 cent as we assume in above Step - II = (x-3) = ax7 + bxy - 12) where a,beIN By (1) 912) CK+1) = ar 7 + b xy + y = ax7+(b +1)xy = (x+1) is formed by wring u cent and 7 cent = PCH+1) is also true. Thus, Den) is true & n7/18. (Proved).