Chapter 2: Problem 51
Assuming that \(1+2+\cdots+n=a n^{2}+b n+c\) find \(a, b,\) and \(c\) by substituting three values for \(n\) and thereby obtaining a system of linear equations in \(a\) \(b,\) and \(c\)
Short Answer
Expert verified
The values are \(a = \frac{1}{2}\), \(b = \frac{1}{2}\), and \(c = 0\).
Step by step solution
01
Understanding the problem
We need to find constants \(a\), \(b\), and \(c\) for which \(1 + 2 + \cdots + n = a n^2 + b n + c\) holds true for all \(n\). To do this, we'll substitute values for \(n\) to create a system of equations involving these constants.
02
Using a known formula
Recall that the sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). We will use this formula to check the sum for specific \(n\) values and compare them to \(a n^2 + b n + c\).
03
Substitute \(n = 1\)
Calculate the sum for \(n = 1\): \(1 = a(1)^2 + b(1) + c = a + b + c\). This gives us the first equation: \(a + b + c = 1\).
04
Substitute \(n = 2\)
Calculate the sum for \(n = 2\): \(1 + 2 = 3 = a(2)^2 + b(2) + c = 4a + 2b + c\). This gives us the second equation: \(4a + 2b + c = 3\).
05
Substitute \(n = 3\)
Calculate the sum for \(n = 3\): \(1 + 2 + 3 = 6 = a(3)^2 + b(3) + c = 9a + 3b + c\). This gives us the third equation: \(9a + 3b + c = 6\).
06
Setting up the system of equations
Now, we have the following system of equations: 1. \(a + b + c = 1\)2. \(4a + 2b + c = 3\)3. \(9a + 3b + c = 6\)
07
Solve the system of equations
Subtract the first equation from the second: \((4a + 2b + c) - (a + b + c) = 3 - 1\)which simplifies to:\(3a + b = 2\). Next, subtract the second equation from the third:\((9a + 3b + c) - (4a + 2b + c) = 6 - 3\)which simplifies to:\(5a + b = 3\). Now we have two new equations:1. \(3a + b = 2\)2. \(5a + b = 3\)
08
Find \(a\) and \(b\)
Subtract the first new equation from the second:\((5a + b) - (3a + b) = 3 - 2\)This simplifies to:\(2a = 1\)So, \(a = \frac{1}{2}\).Plug \(a = \frac{1}{2}\) into \(3a + b = 2\):\(3(\frac{1}{2}) + b = 2\)\(\frac{3}{2} + b = 2\)Subtract \(\frac{3}{2}\) from both sides: \(b = \frac{1}{2}\).
09
Find \(c\)
Using \(a = \frac{1}{2}\) and \(b = \frac{1}{2}\) in the first equation \(a + b + c = 1\): \(\frac{1}{2} + \frac{1}{2} + c = 1\). This simplifies to \(1 + c = 1\). Therefore, \(c = 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
System of Equations
A system of equations consists of a set of multiple equations that share the same variables. In solving a system of linear equations, we aim to find the values of the variables that satisfy all equations in the system. This is fundamental in algebra.
- The equations can be linear or non-linear (though here we are considering linear).
- The solution can be found using various methods like substitution, elimination, or matrix operations.
Constant Terms
In algebra, a constant is a fixed value that does not change. When expressed within an equation, constants appear as standalone numbers, without attached variables. Constant terms may include integers, fractions, or decimals.
- They provide reference points in equations.
- Understanding their role helps in rearranging equations.
Sum of Natural Numbers Formula
The sum of the first \(n\) natural numbers can be calculated using a specific formula: \( \frac{n(n+1)}{2} \). This formula provides a quick method to determine the total of a sequence of numbers starting from 1 up to \(n\).
- Natural numbers are positive integers starting from 1.
- Understanding this formula is crucial in various mathematical applications.