Solving for the Sum of the First n Even Numbers: A Seoer's Perspective
In this article, we delve into the mathematical challenge of finding the sum of the first (n) even numbers, given that the sum of the first (m) odd numbers exceeds it by 212. We'll provide a step-by-step solution, enriching the content with SEO optimization strategies to ensure it gets discovered and indexed by Google.
Introduction
This article aims to solve a mathematical problem that involves the sum of odd and even numbers. The challenge is to find the sum of the first (n) even numbers, given that the sum of the first (m) odd numbers exceeds it by 212. We will explore the formulas for the sums of these sequences and use quadratic equations to find the possible values of (n).
Formulating the Problem
Let's begin by defining the sums of the series we are dealing with.
Sum of the First (m) Odd Numbers
The sum of the first (m) odd numbers is given by the formula:
[S_{text{odd}}(m) m^2]Sum of the First (n) Even Numbers
The sum of the first (n) even numbers is given by the formula:
[S_{text{even}}(n) n(n 1)]Solving the Problem
Given that the sum of the first (m) odd numbers exceeds the sum of the first (n) even numbers by 212, we can set up the following equation:
[m^2 n(n 1) 212]Rearranging this equation gives us:
[m^2 - n(n 1) 212]Expanding the equation:
[m^2 n^2 n 212]We can write (m) in terms of (n) as:
[m^2 n^2 - n 212]Next, we express (m^2) in terms of (n) by setting:
[k m]Thus, the equation becomes:
[k^2 n^2 - n 212]We need to find (k^2) to be a perfect square. Rearranging the equation we have:
[k^2 - n^2 - n - 212 0]Using the quadratic formula for (n):
[n frac{-(-1) pm sqrt{(-1)^2 - 4 cdot 1 cdot (-212 - k^2)}}{2 cdot 1}][n frac{1 pm sqrt{1 848 4k^2}}{2}][n frac{1 pm sqrt{4k^2 847}}{2}]Finding Suitable Values for (n)
To ensure (n) is a positive integer, the expression under the square root must be a perfect square. Thus, we set:
[4k^2 - 847 t^2]This can be factorized as:
[4k^2 - t^2 847]Using the difference of squares, we get:
[(2k - t)(2k t) 847]Next, we need to find the factor pairs of 847. The prime factorization of 847 is:
[847 7 times 121 7 times 11^2]Thus, the factor pairs of 847 are:
[1, 847, 7, 121, 11, 77]We will solve for each factor pair (a, b).
Factor Pair (1, 847)
[2k - t 1, quad 2k t 847]Adding these equations:
[4k 848 implies k 212]Substituting (k 212) into (2k - t 1):
[424 - t 1 implies t 423]Now finding (n):
[4k^2 - 847 423^2 implies n frac{1 423}{2} 211]Factor Pair (7, 121)
[2k - t 7, quad 2k t 121]Adding these equations:
[4k 128 implies k 32]Substituting (k 32) into (2k - t 7):
[64 - t 7 implies t 57]Now finding (n):
[4k^2 - 847 57^2 implies n frac{1 57}{2} 28]Factor Pair (11, 77)
[2k - t 11, quad 2k t 77]Adding these equations:
[4k 88 implies k 22]Substituting (k 22) into (2k - t 11):
[44 - t 11 implies t 33]Now finding (n):
[4k^2 - 847 33^2 implies n frac{1 33}{2} 16]Thus, the possible values of (n) are 211, 28, and 16. The sum of all possible values of (n) is:
[211 28 16 255]Conclusion
The conclusion to this problem is that the sum of all possible values of (n) is:
[boxed{255}]