Sequence problem

  • Thread starter erisedk
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  • #1
erisedk
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Homework Statement


Consider the sequence in the form of groups (1), (2,2), (3,3,3), (4,4,4,4), (5,5,5,5,5),...

i. The 2000th term of the sequence is not divisible by
(A) 3 (B) 9 (C) 7 (D) None of these

ii. The sum of the first 2000 terms is
(A) 84336 (B) 96324 (C) 78466 (D)None of these

iii. The sum of the remaining terms in the group after 2000th term in which 2000th term lies is
(A) 1088 (B) 1008 (C) 1040 (D) None of these

Homework Equations




The Attempt at a Solution


I don't know where to start!
For ii, I need to find 1^2 + 2^2 + 3^2 + 4^2... for which I have the formula n(n+1)(2n+1)/6, but I don't know what the preceding number of the group in which the 2000th term lies will be, so I can't even approximate an answer.
Just please tell me how to begin!
 

Answers and Replies

  • #2
haruspex
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I need a bit more background. In what sense is (5,5,5,5,5) a group, and in what sense might it be divisible by an integer?
 
  • #3
erisedk
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1-first term 2-second term 2-third term 3-fourth term 3-fifth term 3-sixth term 4- seventh term
If the questions were-
The sixth term of the sequence is not divisible by-
(A)1 (B)2 (C)3
Then the answer would be (B) as 3 is not divisible by 2.

Sum of the first six terms-
Ans: 14

The sum of the remaining terms in the group after 6th term in which 6th term lies is
Ans: 0
OR
The sum of the remaining terms in the group after 5th term in which 5th term lies is
Ans: 3
 
  • #4
erisedk
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Sorry for the repeat post.

Moderator note: It's now deleted.
 
Last edited by a moderator:
  • #5
haruspex
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Ok, I see.
Not sure why you think you want the formula for the sum of consecutive squares. You are not trying to sum the series.
How many terms are there before the occurrence of the first n?
 
  • #6
erisedk
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Before the occurrence of the first n??
 
  • #7
haruspex
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Before the occurrence of the first n??
As in, zero before the first 1, one before the first 2, three before the first 3, six before the first 4,... How many before the first n?
 
  • #8
erisedk
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n(n-1)/2 before the first n. What now?
 
  • #9
haruspex
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n(n-1)/2 before the first n. What now?
So if the 2000th term is n, what equation can you write down?
 
  • #10
erisedk
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n(n-1)/2 < = 2000 ?
 
  • #11
erisedk
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Oh ok! n=63 gives 1953 numbers before 63. I got all the answers! Thank you so much :D
 

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