If the sum of all the even numbers from 1 to ( is an odd number) is , what is the value of ?
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Got 139 by setting up sum of evens up to n-1 (since n is odd) equal to 69*70, then solving k(k+1)=69*70. But is there a faster way? Took me a bit to see the pattern.
That's the standard way. Alternatively, recognize that sum of first k evens = k(k+1), so k=69 directly. Since n is odd, the last even is n-1, so n-1=2*69 => n=139.
Nice! I initially tried plugging choices but that was slow. Your method is cleaner.
Why is n odd? The problem says n is an odd number, so the even numbers from 1 to n are 2,4,...,n-1. That's the key. Without that, I'd have been stuck.
Exactly. If n were even, the sum would be different. The odd condition ensures the last even is n-1.
I got 139 but I did it by listing? No, too many. I used formula: sum of evens up to n-1 = 2+4+...+(n-1) = 2(1+2+...+(n-1)/2) = 2 * ((n-1)/2)((n-1)/2+1)/2 = ((n-1)/2)((n-1)/2+1). Set equal to 69*70, so (n-1)/2=69 => n=139. Works.
That's exactly what I did. Good to see confirmation.
This felt more like a 650 level. The trick is seeing that 69*70 is the product of consecutive integers, which matches k(k+1).
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.