Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
Answer:
- n = 11
Step-by-step explanation:
We can observe general terms of n:
- aₙ = n + 1 + 2n + 2 + 3n + 3 + ... + (n - 1)n + (n - 1)
Rewrite it as:
- aₙ = (n + 1) + 2(n + 1) + 3(n + 1) + ... (n - 1)(n + 1) =
- (n + 1)(1 + 2 + 3 + ... + n - 1) =
- (n + 1)(1 + n - 1)(n - 1)/2 =
- (n - 1)n(n + 1)/2
We need to find the least n for which aₙ > 500:
- (n - 1)n(n + 1)/2 > 500 ⇒ (n - 1)n(n + 1) > 1000
If n = 10:
- a₁₀ = 9*10*11 = 990 < 1000
If n = 11:
- a₁₁ = 10*11*12 = 1320 > 1000
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.
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.