i)
Given : u1 = 1, u2 = 7, u(n+1) = 3un + Pn
u2 = 3u1 + P(1) → 7 = 3(1) + P
P = 7 - 3 = 4
u3 = 3u2 + P(2) = 3(7) + 2(4) = 21 + 8 = 29
ii)
Given : un = a(3ⁿ) + bn + c
So,
u1 = a(3¹) + b(1) + c → 1 = 3a + b + c ①
u2 = a(3²) + b(2) + c → 7 = 9a + 2b + c ②
u3 = a(3³) + b(3) + c → 29 = 27a + 3b + c ③
There are now 3 equations with 3 unknowns. You can solve this using GC and still get the full credit/marks, but manually it would be something like :
③ - ② : 22 = 18a + b→ ④
② - ① : 6 = 6a + b →⑤
④ - ⑤ : 16 = 12a → a = 16/12 = 4/3
Sub a = 4/3 into ⑤,
6 = 6(4/3) + b → b = 6 - 8 = -2
Sub a = 4/3, b = -2 into ①,
1 = 3(4/3) - 2 + c → c = 1 - 4 + 2 = -1
∴ a = 4/3, b = -2, c = -1

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