91Ó°ÊÓ

First, we have to find the actual value of the difference given, which is \(57^2 - 13^2\). The formula for the difference of two squares is: \[A^2 - B^2 = (A+B)(A-B)\] Using this formula with \(A = 57\) and \(B = 13\): \[57^2 - 13^2 = (57+13)(57-13)\]

Now, we will find the values of \(57 + 13\) and \(57 - 13\): \(57 + 13 = 70\) \(57 - 13 = 44\)

Now, multiply the values obtained in the previous step: \(70 * 44 = 3080\) Now we know the sum of the consecutive odd positive integers is \(3080\).

The sum of the first n odd positive integers is given by the formula: \[S = n^2\] So, we have to find an odd number which when subtracted from this sum, leaves us with a perfect square. Now, let's try to find such an odd number that when subtracted from \(3080\), it leaves a perfect square: For example, \(3081 - 53 = 3027\) which is not a perfect square. \(3081 - 15 = 3066\) which is not a perfect square. \(3081 - 1 = 3080\) which is not a perfect square. \(3081 - 21 = 3060 \) which is also not a perfect square. However, \(3080 - 45 = 3035\) The result, \(3035\), is not a perfect square, but trying to represent it as the difference of consecutive squares, \(55^2 - 10^2 = 3025\) Then: \(3080 - 45 = 3025 + 5\) So, the sum of consecutive odd positive integers is equal to \(3080\). Thus, the consecutive odd positive integers are: \(3025\), \(3027\), \(3029\), ..., \(3045\) Let's break this sum down into the series: \[5^2 + 7^2 + 11^2 + ... + 45\] \[5^2 + (5^2 + 2^2) + (5^2 + 6^2) + ... + 45\] Now, since there are 5 parts in this sum, we can represent these parts as following: \[5(5^2) + (0^2 + 2^2 + 6^2 + 8^2 + 12^2)\] \[5(25) + (0 + 4 + 36 + 64 + 144)\] \[125 + 248 = 373\] Now we have a sum of 373 with the consecutive odd positive integers starting at 3729 and ending at 3845. Thus, the consecutive odd positive integers with a sum equal to \(57^2 - 13^2\) are: \[3729, 3731, 3733, ..., 3845\]