Answered

Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

2022-01-27
Suppose [tex]a_{n}[/tex] as the sum of the numbers in which the quotient and the remainder are equal when divided by [tex]n.[/tex] For example, [tex]a_{5}=6+12+18+24=60.[/tex] Find the least [tex]n[/tex] such that [tex]a_{n}\ \textgreater \ 500.[/tex]​

Sagot :

mhanifa

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

LammettHash

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.

Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.