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Answer:
Step-by-step explanation:
We can observe general terms of n:
Rewrite it as:
We need to find the least n for which aₙ > 500:
If n = 10:
If n = 11:
So the least n is 11
If I'm understanding the construction of this sequence correctly, we have
• a₁ = 0
(the sum is empty since there is no positive integer b such that b•1 + b = 1)
• a₂ = 3
(since 3 = 1•2 + 1)
• a₃ = 4 + 8 = 12
(since 4 = 1•3 + 1 and 8 = 2•3 + 2)
• a₄ = 5 + 10 + 15 = 30
(since 5 = 1•4 + 1, 10 = 2•4 + 2, and 15 = 3•4 + 3)
and so on.
Notice that for n ≥ 2, aₙ is simply the sum of the first n - 1 multiples of n + 1. So
[tex]\displaystyle a_n = \sum_{i = 1}^{n - 1} (n+1)i[/tex]
Recall that
[tex]\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}2[/tex]
Then
[tex]\displaystyle a_n = (n+1) \sum_{i=1}^{n-1} i = (n+1) \cdot \frac{(n-1)n}2 = \frac{n(n^2-1)}2[/tex]
Solve for n such that aₙ > 500 :
n (n² - 1)/2 > 500
n (n² - 1) > 1000
n³ - n > 1000
We can solve this by inspection. Noticing that 10³ = 1000, if we replace n = 10 we get
10³ - 10 = 990 > 1000
which is false, but the difference is quite small. So we move up to n = 11 and find
11³ - 11 = 1320 > 1000
which is true, so n = 11 is the least number such that aₙ > 500.