If a1 = 1, a2 = 1 and an = 2an - 1 + an - 2 n ≥ 3, n ∈ N, then find the first six terms of the sequence.
Solution :
First and second terms are 1.
an = 2an - 1 + an - 2
n = 3
an = 2an - 1 + an - 2
a3 = 2a3-1 + a3-2
a3 = 2a2 + a1
a3 = 2(1) + 1
a3 = 3
|
n = 4
an = 2an - 1 + an - 2
a4 = 2a4-1 + a4-2
a4 = 2a3 + a2
a4 = 2(3) + 1
a4 = 7
|
n = 5
an = 2an - 1 + an - 2
a5 = 2a5-1 + a5-2
a5 = 2a4 + a3
a5 = 2(7) + 3
a5 = 17
|
n = 6
an = 2an - 1 + an - 2
a6 = 2a6-1 + a6-2
a6 = 2a5 + a4
a5 = 2(17) + 7
a5 = 41
|
Hence the first 6 terms are 1, 1, 3, 7, 17, 41.
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