Question 1209312: In the sequence 1, 3, 6, 10, 15, 21, ..., in the first 1000 terms of the sequence, how many terms are multiples of 5?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52752)   (Show Source):
You can put this solution on YOUR website! .
In the sequence 1, 3, 6, 10, 15, 21, ..., in the first 1000 terms of the sequence,
how many terms are multiples of 5?
~~~~~~~~~~~~~~~~~~~~~~~
The n-th term of this sequence,  , is the sum of the first n natural numbers
 = 1,
 = 3 = 1 + 2,
 = 6 = 1 + 2 + 3,
 = 10 = 1 + 2 + 3 + 4,
 = 15 = 1 + 2 + 3 + 4 + 5,
and so on. For the sum of the first natural numbers, there is the formula
=  .
So, the question is: how many terms of the first 1000 terms are multiples of 5.
In order for the term be a multiple of 5, EITHER n OR (n+1) must be multiple of 5
(here "either - or" is exclusive).
Of the first 1000 numbers n = 1, 2, 3, 4, 5, . . . , 1000, there are 1000/5 = 200 multiple of 5.
Of the first 1000 numbers n+1 = 2, 3, 4, 5, . . . , 1001, there are  = 200 other numbers multiple of 5.
Thus, in all, among first 1000 numbers =  , there are 200 + 200 = 400 numbers multiple of 5.
ANSWER. In the sequence 1, 3, 6, 10, 15, 21, ..., in the first 1000 terms of the sequence,
400 numbers are multiple of 5.
Solved.
Answer by greenestamps(13195)  (Show Source):
You can put this solution on YOUR website!
ANSWER: 400 (2/5 of the numbers in the sequence)
-----------------------------------------------------------------------
(1) By looking for a pattern in the numbers themselves....
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, ...
Observe that 2 numbers in each sequence of 5 are multiples of 5.
-----------------------------------------------------------------------
(2) By looking at the sequence of numbers modulo 5....
1, 3, 1, 0, 0, 1, 3, 1, 0, 0, 1, 3, 1, 0, 0, ...
The number is a multiple of 5 if and only if the number modulo 5 is 0.
-----------------------------------------------------------------------
(3) By analyzing the formula that produces the terms of the sequence....
The numbers in the sequence are the triangular numbers -- the n-th number in the sequence is the sum of the integers from 1 to n. The formula for the n-th triangular number is
That number will be a multiple of 5 if an only if either n or (n+1) is a multiple of 5.
In any sequence of 5 consecutive integers, exactly 1 of them will have n being a multiple of 5 and exactly 1 of them will have (n+1) being a multiple of 5. So in any long string of consecutive integers, 2/5 of them will produce numbers in the sequence that are multiples of 5.
|
|
|
