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A sequence is defined by a_1 = 1 and a_{n+1} = 3 a_n + 2^n for all n…

Question

A sequence is defined by a_1 = 1 and a_{n+1} = 3 a_n + 2^n for all n ≥ 1. Find a_8.

✓ Verified answer: 6305checked by our engine — not a guess

Step-by-step solution

Solve the linear recurrence a_{n+1} = 3a_n + 2^n.
The homogeneous solution is C·3^n.
For a particular solution try a_n = D·2^n: D·2^{n+1} = 3D·2^n + 2^n ⇒ 2D = 3D + 1 ⇒ D = -1.
So a_n = C·3^n - 2^n.
Using a_1 = 1: 3C - 2 = 1 ⇒ C = 1.
Hence a_n = 3^n - 2^n.
Then a_8 = 3^8 - 2^8 = 6561 - 256 = 6305.

Final answer6305

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