In a sequence of positive integers, every odd-numbered term after the first term is 1 more than twice the previous term and every even-numbered term starting from the 2nd term is 2 less than 3 times the previous term. If the 2nd term of the sequence is 1, then how many terms out of the first 20 terms are even?
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Wait, I'm confused about the wording. Is the rule based on the position of the term or the value of the term?
It's based on the position — odd-numbered terms vs even-numbered terms.
Yeah, position. That tripped me up too until I re-read it.
This one took me way longer than it should have. I started listing terms and then realized I should just check the parity pattern. But it's weird how it flips back and forth.
If the 2nd term is 1, then the 3rd term is 2*1+1=3, and the 4th is 3*3-2=7. Then 5th is 2*7+1=15, 6th is 3*15-2=43. So starting from 3 onward it looks like all odds? Let me double-check... 15 odd, 43 odd, 87 odd... yeah seems like after the 2nd it's always odd. So only the 2nd term is even?
Don't spoil it! But yeah, I got the same pattern.
Careful — you should verify the parity claim with the actual rules, not just the first few.
Is there a faster way than computing all 20 terms? I feel like there's a parity shortcut here.
Think about parity of 2x+1 and 3x-2 mod 2.
If a term is odd, 3x-2 is odd*3-2 = odd-even = odd. And 2x+1 is always odd. So once odd, always odd.
What it tests
Your ability to spot and extend arithmetic and geometric sequences and work with recursive definitions.
Common trap
Off-by-one errors in indexing (a₀ vs a₁) or confusing an arithmetic difference with a geometric ratio.