Solve these recurrence relations together with the initial conditions given. a) an = an-1 + 6an-2 for n ≥ 2, a0 = 3, a1 = 6
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Let's try to find a formula for an in terms of n: an = an−1 +6an−2 an−1 = an−2 +6an−3 Substituting the second equation into the first, we get: an = (an−2 +6an−3) +6an−2 an = 7an−2 +6an−3 We can continue this process to get: an = 13an−3 +18an−4 This suggests Show more…
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