Question

Problem 4 (25 points + 5 points bonus (total 30)). Suppose do = 0, a1 = 2, and an+2 = 4an+1 - 4an + n2 – 5n + 2. Show that n divides an for all n > 1. Note: this is a very difficult problem—more difficult than the other problems, which is why there are 5 bonus points associated with it. You may wish to begin by re-indexing, so that do = a1 = 0, and you may have to differentiate the appropriate power series to solve this problem. Additionally, it might help you spot the pattern if you first compute out the first several terms of the series. Solution. Collaborators: No one.

Answer 1

- 47 0 Given, - Ontz 4 anri-4an tn? -5n+2; a 0, 91.2 This can be written as, antz-4 ante 44 an = n²5n72- The corresponding homogeneous part is ... ant2-4ant than to The auxiliary equation is: . r²- 40+450 . - rot (r-.27²50 care 2,2 The solution is ang = (A + nB 7217 Now, we have, = n 5n.t2. so our trail solution is Taz cn?tonite! . say, ans (n²t Dritt antis {n+12+(2+1) TE cm2 + m +1) + D(n+1) + s imet (2C++)n + +D+E) Antz= c (n+2)? + D(n+2) te : :ċ (42+46 +4) + D(n+22+€ ccm? + 4C+Dn + f+20+0)

plugging in o me have, cn't Kc+o)n + (46 +20+€) - 4 (6m2. +626.toln + (c+0E)) + 4 [cm2 ton +[] = m2–5n+2 -) (-46 +45)m? + [4C+D-4(2C+D) +up]n + 46+20+E-46-40-4€ +4€ an²-5272 a L @ D-46 -5 cn2+ (acto-8(-45 +46) at 6207 € ) = 72-5672 cn²ė (0-40) int (€-20%=n? - snt2 g Call & [D-463-5 Da -5th = -1 .10-1) : Ė – 2D = 2 = TESO => rats m2-n] Thus, ans and at. an= (A + 2B) 2" + n²_n] - o . me have ao= o . and a=2 a B 2" to picons 2 BE 2!m?]. :

Plugging in ③ ; . . an= n2n. zn 7. n²_n = 2n + na-n = n +n2 ans ntn? More over, an can be written as . anantn²= n(n+1. - . n . divides an..