Did you know that \(1225\) can be written as the sum of seven consecutive integers?
That is, \[1225=172+173+174+175+176+177+178\]
The notation below illustrates a mathematical short form used for writing the above sum. This notation is called Sigma Notation.
\[\sum_{i=172}^{178}i = 1225\]
How many ways can the number \(1225\) be expressed as the sum of an even number of consecutive positive integers?
Suppose \(k\) is even. We can write \(k\) consecutive integers as \[n-\left(\frac{k}{2}-1\right), \ \ldots,\ n,\ n+1,\ \ldots,\ n +\left(\frac{k}{2}-1\right),\ n+\frac{k}{2}\]
Here, \(n\) and \(n+1\) are the middle numbers in the sum, and there are \(\frac{k}{2}-1\) integers less than \(n\) in the sum and \(\frac{k}{2}\) integers greater than \(n\) in the sum.
We can write the sum of these integers in this way: \[\left(n-\left(\frac{k}{2}-1\right)\right)+ \cdots +n + (n+1) + \cdots + \left(n+\left(\frac{k}{2}-1\right)\right) + \left(n+\frac{k}{2}\right)\]
This simplifies to \(kn + \frac{k}{2}.\)
For example, four consecutive integers can be expressed as \(n-1\), \(n\), \(n+1\), and \(n+2\), where \(n\) is an integer.
Their sum is \((n-1) + n + (n+1)+(n+2)=4n +
2.\)
Notice that \(kn + \frac{k}{2} = k(n+\frac{1}{2}).\) Thus, if this sum is equal to \(1225\), then \(k(n+\frac{1}{2})=1225.\) Multiplying both sides by \(2\), we have \[\begin{aligned} 2k\left(n+\frac{1}{2}\right) &=2(1225)\\ k(2n+1) & = 2450 \end{aligned}\] Since \(n\) is an integer, then \(2n+1\) is an odd integer. Therefore, we’re looking for factor pairs of \(2450\), where one factor is even and the other is odd.
Since \(2450=2(5^2)(7^2)\), the positive odd divisors of \(2450\) are \(1\), \(5\), \(7\), \(25\), \(35\), \(49\), \(175\), \(245\) and \(1225.\)
For each odd divisor, \(2n+1\), of \(2450\), we determine \(n\) and \(k= \frac{2450}{2n+1}.\) The \(k\) integers that sum to \(1225\) will then be \(n-\left(\frac{k}{2}-1\right)\), \(\ldots\), \(n\), \(n+1\), \(\ldots\), \(n +\left(\frac{k}{2}-1\right)\), \(n+\frac{k}{2}.\) This is summarized in the table below.
| Odd Divisor (\(2n+1\)) | \(n\) | Number of integers (\(k\)) | Sum of Integers |
|---|---|---|---|
| \(1\) | \(0\) | \(2450\) | \((-1224) + (-1223) + \cdots +0 + \cdots + 1224 + 1225\) |
| \(5\) | \(2\) | \(490\) | \((-242)+(-241)+\cdots +2+\cdots + 246+247\) |
| \(7\) | \(3\) | \(350\) | \((-171)+(-170)+\cdots+3+\cdots+177+178\) |
| \(25\) | \(12\) | \(98\) | \((-36)+(-35)+\cdots+12+\cdots+60+61\) |
| \(35\) | \(17\) | \(70\) | \((-17)+(-16)+\cdots+17+\cdots+51+52\) |
| \(49\) | \(24\) | \(50\) | \(0+1+\cdots+24+\cdots+48+49\) |
| \(175\) | \(87\) | \(14\) | \(81+82+83+84+85+86+87+88+\space\)\(89+90+91+92+93+94\) |
| \(245\) | \(122\) | \(10\) | \(118+119+120+121+122+123+\space\)\(124+125+126+127\) |
| \(1225\) | \(612\) | \(2\) | \(612+613\) |
For \(k=14\), \(10\), and \(2\), all integers in the sum are positive.
Thus, there are three ways to express \(1225\) as the sum of an even number of consecutive positive integers.
Extension: Determine the number of ways the number \(1225\) can be expressed as the sum of an odd number of consecutive positive integers.