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Sums of consecutive integers?
1. (Easy) Find a set of two consecutive integers whose sum is equal to the next integer immediately following them.
2. Find a set of three consecutive integers whose sum is equal to the sum of the next two consecutive integers immediately following them.
3. Find a set of four consecutive integers whose sum is equal to the sum of the next three consecutive integers immediately following them.
4. Given any n > 1, is it always possible to find a set of n consecutive integers whose sum is equal to the sum of the next n - 1 consecutive integers immediately following them?
When such a sequence of integers exists for some n, where does the sequence begin, i.e. what is the smallest integer in the first sum?
3 Answers
- Math ChickLv 41 decade agoFavorite Answer
smallest such integer is (n - 1)^2 for n > 1.
Source(s): simple algebra - 1 decade ago
1. 1+2 = 3
2. 4+5+6 = 7+8 (15 = 15)
3. 9+10+11+12 = 13+14+15 (42 = 42)
(Not asked for, but:
16+17+18+19+20 = 21+22+23+24 (90 = 90)
25+26+27+28+29+30 = 31+32+33+34+35 (165 = 165))
4. Yes.
I began by asking what the average term is in each series. We start with 3 numbers that sum to 6 (1+2 = 3, grand total is 6), then 5 that sum to 30, 7 that sum to 84, 9 that sum to 180, and 11 that sum to 330.
Notice that there is always an odd number of terms in each sequence (n + n -1 is always the sum of an odd and an even number, and an odd plus an even is odd).
The average of each sequence is thus 6/3, 30/5, 84/7, 180/9, and 330/11, or 2, 6, 12, 20, 30.
The difference between the terms of this sequence (the deltas) is even more interesting. If we start with zero, to get the terms of the sequence we must add 2, 4, 6, 8, 10, ....
Thus, the next number in the sequence is 30 + 12, or 42, and the series with 13 numbers (7 consecutive and the 6 consecutive following summing to the same value) would have as its average value 42, its grand total 42*13 = 546, and the sum of each side 273.
Since 42 is the average value and the series is an odd number of consecutive integers, 42 is also the median value and the last number of the left-hand side. The sequence begins at 42 - (n - 1) and ends at 42 + (n - 1).
In this case n is 7, n - 1 is 6, so the first sequence goes from 36-42 and the second from 43-48.
36+37+38+39+40+41+42 = 43+44+45+46+47+48 (273 = 273)
The sequence above is known as A002378 (found via a search), and is apparently known as "Oblong (or pronic, or heteromecic) numbers: n(n+1)." Although as you've defined things above, it would be more appropriate to call it (n-1)n.
For any n you can find the average, and thus the median of the overall sequence, by calculating (n-1)n and then subtracting (n-1) to find the starting point for the first sum.
- 1 decade ago
1) (x) + (x +1) = x +2
2) (x ) + (x+1) + (x+2) = (x+3) + (x +4)
3) (x) + (x + 1) + (x +2) + (x +3) = (x+4) + (x+5)+ (x+6)
