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Magazine Chapter 2: Problem 10
Prove that the sum of \(n\) squares can be found as follows $$ 1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6} $$
Short Answer
Expert verified
By verifying the base case where the formula holds for n=1 and using the principle of mathematical induction to prove the formula is true for n = k+1, assuming it is true for n=k, the sum of squares formula is proved to hold for all natural numbers n.
Step by step solution
01
Understand the Problem
The task is to prove that the sum of the squares of the first n natural numbers is given by the formula \(\frac{n(n+1)(2n+1)}{6}\). This is a mathematical induction problem, where we need to show that the statement holds for some initial natural number, usually 1, and then assume it holds for an arbitrary natural number k, and prove it for the next natural number, k+1.
02
Verify the Base Case
We start by checking the base case to ensure the validity of the formula for the first natural number, which is 1. Plug n=1 into the given formula:\[1^2 = \frac{1(1+1)(2(1)+1)}{6} = \frac{1(2)(3)}{6} = 1\] Since the formula is true for n=1, we can proceed with the inductive step.
03
Inductive Hypothesis
Assume that the formula holds true for a particular natural number k, such that:\[1^2 + 2^2 + 3^2 + \ldots + k^2 = \frac{k(k+1)(2k+1)}{6}\] This assumption is called the inductive hypothesis. Our goal now is to show that the formula also holds true for k+1.
04
Prove for the Next Case (k+1)
Starting from the inductive hypothesis, we need to show that adding the square of k+1 to the sum maintains the formula's validity:\[1^2 + 2^2 + 3^2 + \ldots + k^2 + (k+1)^2 = \frac{(k+1)(k+2)(2(k+1)+1)}{6}\] Simplify and add \( (k+1)^2 \) to the left side of the inductive hypothesis:\[\frac{k(k+1)(2k+1)}{6} + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6}\] Combine the terms over the common denominator:\[= \frac{k(k+1)(2k+1) + 6(k+1)^2}{6}\] Now we expand and simplify the numerator to match the target form.
05
Simplify the Expression
Expanding the numerator yields:\[k(k+1)(2k+1) + 6(k+1)^2 = k(2k^2 + 7k + 6) + 6(k+1)\] Factor out a \(k+1\) to simplify the equation further:\[= (k+1)(2k^2 + 7k + 6 + 6)\] Simplify inside the brackets:\[= (k+1)(2k^2+7k+12)\] Factor the quadratic expression in the brackets to find the product of three consecutive integers:\[= (k+1)(k+2)(2k+3)\] This simplication is equal to the right side of our equation for k+1.
06
Conclusion
Since we have shown that adding \( (k+1)^2 \) to the sum \( 1^2 + 2^2 + \ldots + k^2 \) results in the sum \( 1^2 + 2^2 + \ldots + k^2 + (k+1)^2 \) and it fits the formula \( \frac{(k+1)(k+2)(2k+3)}{6} \) just as the formula predicts, this completes the inductive step. Therefore, by mathematical induction, the given formula holds true for all natural numbers n.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sum of Squares Formula
The sum of squares formula is a fundamental algebraic expression in mathematics, particularly useful in various fields like statistics, physics, and computer science. It provides a concise way to express the cumulative total of each natural number up to the specified limit squared.
For example, the sum of the squares of the first two natural numbers is calculated as: \(1^2 + 2^2 = 1 + 4 = 5\). However, computing this sum manually becomes impractical for larger numbers. The sum of squares formula equips us with a shorthand to compute this quickly and efficiently:\[1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}\]
This elegant expression circumvents the need for calculating each square and adding them sequentially, streamlining the process especially for large values of n.
For example, the sum of the squares of the first two natural numbers is calculated as: \(1^2 + 2^2 = 1 + 4 = 5\). However, computing this sum manually becomes impractical for larger numbers. The sum of squares formula equips us with a shorthand to compute this quickly and efficiently:\[1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}\]
This elegant expression circumvents the need for calculating each square and adding them sequentially, streamlining the process especially for large values of n.
Proof by Induction
Proof by induction is a powerful mathematical technique used to establish the truth of an infinite number of statements. When dealing with natural numbers, induction is a logical progression showing that if a statement holds true for the initial case (usually for n=1), and if we can show that the truth of the statement for an arbitrary number implies its truth for the next number, then the statement must hold for all natural numbers.
In any proof by induction, the process entails two important steps: the base case and the inductive step. The base case proves the statement for the starting natural number, usually 1. The inductive step involves assuming the statement is true for some natural number k, termed the inductive hypothesis, and then proving it for k+1. If both steps are successfully demonstrated, the original statement is proven true universally for all natural numbers.
In any proof by induction, the process entails two important steps: the base case and the inductive step. The base case proves the statement for the starting natural number, usually 1. The inductive step involves assuming the statement is true for some natural number k, termed the inductive hypothesis, and then proving it for k+1. If both steps are successfully demonstrated, the original statement is proven true universally for all natural numbers.
Natural Numbers
Natural numbers originate from basic counting and are the building blocks in mathematics to express quantity and order. They are the set of positive integers, typically beginning with 1, then 2, 3, 4, and so on indefinitely. We denote the set of natural numbers by the symbol \(\mathbb{N}\).
Natural numbers are not just crucial for counting but also serve as a foundation for more complex structures in mathematics, such as rational numbers, algebra, and number theory. Proving properties involving natural numbers, like the sum of squares formula, often employ the principle of mathematical induction due to the sequential nature of these numbers.
Natural numbers are not just crucial for counting but also serve as a foundation for more complex structures in mathematics, such as rational numbers, algebra, and number theory. Proving properties involving natural numbers, like the sum of squares formula, often employ the principle of mathematical induction due to the sequential nature of these numbers.
Algebraic Manipulation
Algebraic manipulation is the process of reshaping algebraic expressions into more useful or simplified forms using a variety of arithmetic operations and properties such as distributive, associative, and commutative laws. Mastering algebraic manipulation is paramount for solving equations and understanding higher mathematics.
In the context of our problem, algebraic manipulation was essential in simplifying the expression obtained in the inductive step. Starting from the expanded form of the sum of the first k natural numbers squared, we added the subsequent term \((k+1)^2\) and through clever grouping, factoring, and simplifying, we demonstrated that the result matches the general formula for the sum of squares for the case of k+1. This skillful algebraic manipulation is a keystone in successful proof by mathematical induction.
In the context of our problem, algebraic manipulation was essential in simplifying the expression obtained in the inductive step. Starting from the expanded form of the sum of the first k natural numbers squared, we added the subsequent term \((k+1)^2\) and through clever grouping, factoring, and simplifying, we demonstrated that the result matches the general formula for the sum of squares for the case of k+1. This skillful algebraic manipulation is a keystone in successful proof by mathematical induction.
