A recurrence can look almost geometric except for an added constant. A carefully chosen shift can remove that constant. The method is worth understanding because it explains the formula instead of guessing it from a short list of terms.
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Generate enough terms to check your work
Let a1 = 1 and an+1 = 2an + 3 for integers n ≥ 1. The first five terms are 1, 5, 13, 29 and 61.
These terms are not a geometric sequence: the ratios are not constant. Repeated substitution will find a small-index term, but a transformation can reveal the general structure.
Choose a shift that absorbs the constant
Define bn = an + 3. Then bn+1 = an+1 + 3 = 2an + 6 = 2bn.
The new sequence is geometric, with first term b1 = 4 and common ratio 2. Hence bn = 4 × 2n-1, and an = 4 × 2n-1 - 3.
Check the initial value and recurrence
At n = 1, the formula gives 4 - 3 = 1. At n = 5, it gives 64 - 3 = 61. Both agree with the direct calculation.
For a full structural check, substituting the formula into 2an + 3 gives 8 × 2n-1 - 3, which is the formula for an+1. Checking only a few numerical terms would not establish that identity for every n.
Understand how the shift was found
If bn = an + c, requiring bn+1 = 2bn gives 3 + c = 2c, so c = 3. The shift is chosen to make the algebra work.
Keep the starting index visible: a sequence beginning at n = 0 would use a different initial-value formula. This original exercise develops one algebra technique for AMC 12 preparation; it is not an official problem or a prediction of the paper.
Continue reading: AMC 12 Trigonometry: Solve Equations Without Losing Solutions.
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Official information and further reading: Source 1
