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Chapter 23: Problem 710

Prove by mathematical induction \(1^{2}+2^{2}+3^{2}+\ldots+n^{2}=(1 / 6) n(n+1)(2 n+1)\).

Short Answer

Expert verified
By mathematical induction, we proved that the formula for the sum of the first n squares is true for all positive integers n: \(1^2+2^2+3^2+\ldots+n^2 = \frac{1}{6} n(n+1)(2n+1)\). The proof consists of the following steps: 1. Verify the base case (n=1). 2. State the induction hypothesis for n=k. 3. Prove the induction step for n=k+1. 4. Simplify the RHS of the equation. 5. Relate the simplified RHS to the desired form. 6. Conclude that the formula holds true for all positive integers n.

Step by step solution

01

Verify the Base Case

Start by verifying the formula for the base case, n = 1. The left-hand side (LHS) of the equation is the sum of the first n squares, and the right-hand side (RHS) is the given formula. For n = 1: LHS: \(1^2 = 1\) RHS: \(\frac{1}{6}(1)(1+1)(2(1)+1) = \frac{1}{6}(1)(2)(3) = 1\) Since the LHS and RHS are equal for n = 1, the base case is true.
02

State the Induction Hypothesis

Assume that the given formula holds true for n = k, where k is an arbitrary positive integer: \[1^{2}+2^{2}+3^{2}+\ldots+k^{2} = \frac{1}{6}k(k+1)(2k+1).\]
03

Prove the Induction Step

We have to prove that the formula also holds true for n = k+1. Consider the sum of the first (k+1) squares: \(1^2+2^2+3^2+\ldots+k^2+(k+1)^2\) Now, we can apply the induction hypothesis to the sum of the first k squares: \[\biggl(1^{2}+2^{2}+3^{2}+\ldots+k^{2}\biggr) + (k+1)^{2} = \frac{1}{6}k(k+1)(2k+1) + (k+1)^2.\] Next, we need to simplify the RHS of the equation and try to bring it to the desired form for n = k+1:
04

Simplify the RHS

To simplify the RHS, start by factoring out (k+1) from the two terms: \[\frac{1}{6}k(k+1)(2k+1) + (k+1)^2 = (k+1)\biggl(\frac{1}{6}k(2k+1) + (k+1)\biggr).\] Now, focus on simplifying the expression inside the parentheses: \[\frac{1}{6}k(2k+1) + (k+1) = \frac{1}{6}(2k^2 + k) + (k+1).\] Observe that we can rewrite the (k+1) term as \(\frac{6(k+1)}{6}\), then combine the terms under a common denominator: \[\frac{1}{6}(2k^2 + k) + \frac{6(k+1)}{6} = \frac{1}{6}(2k^2 + k + 6k + 6) = \frac{1}{6}(2k^2 + 7k + 6).\]
05

Relate the Simplified RHS to the Desired Form

Notice that we can factor the simplified RHS expression: \[\frac{1}{6}(2k^2 + 7k + 6) = \frac{1}{6}(k+1)((2k+1)+(k+2)).\] Comparing this to the desired form for n = k+1: \[\frac{1}{6}(k+1)(2(k+1)+1)(k+1+1) = \frac{1}{6}(k+1)(2k+3)(k+2).\] Since the simplified RHS is equal to the desired form, we have proven that the given formula holds true for n=k+1, thus completing the induction step.
06

Conclusion

By induction, we have proven that the given formula for the sum of the first n squares is true for all positive integers n: \[1^{2}+2^{2}+3^{2}+\ldots+n^{2} = \frac{1}{6} n(n+1)(2n+1).\]

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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 mathematical expression that provides a shortcut to calculate the sum of squares of the first n natural numbers. Rather than adding each squared number individually, we can use the formula:
  • \(1^{2}+2^{2}+3^{2}+bsp\ldots+n^{2} = \frac{1}{6} n(n+1)(2n+1)\)
This expression simplifies calculations significantly, especially for larger values of n. For instance, if you want to find the sum of squares from 1 to 100, manually computing each square would be tedious, but using the formula, you just need to plug in the number n=100 to get your answer. The formula leverages the symmetrical properties of quadratic numbers and is both efficient and elegant.
Proof by Induction
Proof by induction is a powerful technique used to establish the truth of an infinite sequence of mathematical statements. It is particularly useful for proving formulas like the sum of squares. The process involves two main steps:
  • The base case, where you verify that the formula holds true for the initial value (usually n=1).
  • The inductive step, where you show that if the formula holds for an arbitrary positive integer n=k, then it must also hold for n=k+1.
Induction resembles a domino effect鈥攊f you can knock over the first domino and guarantee that each domino knocks over the next, all the dominoes will fall. Similarly, by establishing these steps, you prove the formula holds for all natural numbers.
Base Case Verification
Base case verification is the first crucial step in proof by induction. It ensures that the formula we are trying to prove is valid for the starting point of the induction process. For the sum of squares formula, we check it for n=1:
  • Left-hand side (LHS): \(1^2 = 1\)
  • Right-hand side (RHS): \(\frac{1}{6}(1)(1+1)(2 \times 1+1) = \frac{1}{6}\times1\times2\times3 = 1\)
Both sides are equal, confirming that the formula is true for n=1. This acts as our starting anchor, showing the formula can at least begin correctly. Without successfully proving the base case, induction cannot proceed, as it forms the foundation for the rest of the proof.
Inductive Step
The inductive step is where the heart of proof by induction lies. During this step, we prove that if the formula is true for n=k, it must be true for n=k+1. Assuming the formula
  • \(1^{2}+2^{2}+3^{2}+bsp\ldots+k^{2} = \frac{1}{6}k(k+1)(2k+1)\)
By induction hypothesis, it holds for n=k. The goal is to show
  • \(1^{2}+2^{2}+3^{2}+bsp\ldots+k^{2}+(k+1)^2 = \frac{1}{6}(k+1)(k+1+1)(2(k+1)+1)\)
We add \((k+1)^2\) to both sides and simplify:
  • \((1^{2}+2^{2}+3^{2}+bsp\ldots+k^{2}) + (k+1)^2 = \frac{1}{6}k(k+1)(2k+1) + (k+1)^2\)
This results in a new expression that matches the formula for n=k+1. This stage completes the proof by establishing that if the formula works for an arbitrary case, it works for the next case, assuring the formula holds for all natural numbers.

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Most popular questions from this chapter

Prove by mathematical induction that the sum of n terms of an arithmetic progression \(\mathrm{a}, \mathrm{a}+\mathrm{d}, \mathrm{a}+2 \mathrm{~d}, \ldots\) is \((\mathrm{n} / 2)[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d})]\), that is \(a+(a+d)+(a+2 d)+\ldots+[a+(n-1) d]=(n / 2)[2 a+(n-1) d]\).

Prove by mathematical induction that, for all positive integral values of \(\mathrm{n}\). $$ 1+2+3+\ldots+n=[n(n+1)] / 2 $$

Let \(\mathrm{x}\) be any real number. Show that \(|\sin \mathrm{n} \mathrm{x}| \leq \mathrm{n}|\sin \mathrm{x}|\) for every positive integer \(\mathrm{n}\).

Prove that the sum of the cubes of the first \(\mathrm{n}\) natural numbers is equal to \(\\{[\mathrm{n}(\mathrm{n}+1)] / 2\\}^{2}\).

Prove by mathematical induction that \([5 /(1 \cdot 2 \cdot 3)]+[6 /(2 \cdot 3 \cdot 4)]+[7 /(3 \cdot 4 \cdot 5)]+\ldots\) \(+\\{(\mathrm{n}+4) /[\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)]\\}\) \(=[n(3 n+7)] /[2(n+1)(n+2)]\)
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