Question

Please answer ALL parts and please show ALL work.

1. Prove that if 5 | na then 5 n. 2. Use mathematical induction to prove that 8i(31 – 3) = (2n - 2)4n(n + 1). 3. Let a1 = 29, 02 = 103 and for n > 3, an = 7an-1 – 10an-2. Find a closed formula for an. (Show your work).

Answer 1

1) let 5 Xn² - Claim- 5xn On a contrary, we assume - n= 5K KEZI - n² = (5K)² = 25 K since 5/25k ² | =) ) | nº. which is a contradiction to 5 Xn² - Herice sfn. RutP (n): 3 81[31-3) = ( 21 -2) 40 (n+1) Base step- for n=1 = 81 3-3) RHS = Ruse (211)-2) 400 (141) = 014)(2) = 0 for n=1, Pln) holos Hypothesis - Assume pln) is tome for n=kie ti (31-3) = (@K-2) 4R (K+1)

To show _ (K+1) is tuue le 81 / 31-3) = [ 2/K+1)-2] 4187 Z Consider & 8; (37-3) 3 8' (3-3)+ plk+) (3*+1/3) -1. = PIK #1) -2] (4x) (x+1) 7 _ $K+13 (36K+1)-3) = (QK-2) 4(K41) K + 4 (k+])a (3K+3-3) = 41K+[K | 2K-22 + 2 (3*)] 541K+j) [ak? -2X+6x] = 41K+) [ 2 k" +48] 5.41K+1) (@k) (k+2) = 4k+1)[Q[k+1)-2 ] [ k*1+1] =) For n=kt lip(K+1) s true Hence by induction we have pecored that 3831-3) =(26-2) 4n (nel)

3) Assume a solution of the from - anaan =) gives an = 7 ani - 1 Dan-2 an = 7ent-1000-2 > - e²- 7r+10=0 l'ei-s) 14-2)=0 - -) eu= 2, & 2 =5 io S solution a n = - is given by 'ci. Je, nit a lesy - 92 '+ c25" where a & G are constante Now, a=29 *%)'q=29= Ġ 23+ 562 - [26+ sej: 29 a, = D 3 = 2^+ , sº = 4 Gt 256 0 by a) = - Tug +25c2=103 , 24 = 29-562 2129-562) 3512= 103 '58 + 15 c2 = 103 = 15c2=45 = G =3 =) 24-29 - 5(3/

20= 14 =) a=7 is fan = 712")+ 3(59) | n21