Let P(n) be the statement that 1^2+2^2+⋯+n^2= n(n+1)(2 n+1)...

Let $P(n)$ be the statement that $1^{2}+2^{2}+\cdots+n^{2}=$ $n(n+1)(2 n+1) / 6$ for the positive integer $n .$ a) What is the statement $P(1)$ ? b) Show that $P(1)$ is true, completing the basis step of the proof. c) What is the inductive hypothesis? d) What do you need to prove in the inductive step? e) Complete the inductive step, identifying where you use the inductive hypothesis.

Let $P(n)$ be the statement that $1^{2}+2^{2}+\cdots+n^{2}=$ $n(n+1)(2 n+1) / 6$ for the positive integer $n .$
a) What is the statement $P(1)$ ?
b) Show that $P(1)$ is true, completing the basis step of the proof.
c) What is the inductive hypothesis?
d) What do you need to prove in the inductive step?
e) Complete the inductive step, identifying where you use the inductive hypothesis.

Key Concepts

Mathematical Induction

A proof technique used to establish the truth of an infinite sequence of statements. It typically involves two main steps: proving the statement for a base case, and then proving that if the statement holds for an integer n, it must also hold for n+1. This method is essential in demonstrating formulas and properties that are defined recursively or over the natural numbers.

Base Case

The starting point of an inductive proof, where the statement is verified for the initial value (commonly n=1). This step ensures that the property holds at the beginning of the sequence, forming the foundation for the inductive process.

Inductive Hypothesis

The assumption that the statement or formula holds for an arbitrary positive integer n. This hypothesis is a critical part of the inductive proof, serving as the given condition from which one proves the truth for n+1.

Inductive Step

The part of an inductive proof where one shows that if the statement holds for an arbitrary integer n (as assumed in the inductive hypothesis), then it also holds for n+1. This step bridges the truth of the statement from one case to the next, completing the logical chain from the base case onward.

Summation Formula

A closed-form expression that represents the sum of a sequence of numbers. In the context of the problem, it refers to the formula for the sum of the squares of the first n natural numbers. Such formulas are often proved using mathematical induction to verify their validity for all positive integers.

Transcript

00:01 So we're given this statement and it's set up so we can perform induction.

00:05 So first it asks us to write out what p of 1 is.

00:08 So that's going to be 1 squared which is equal to 1 times 1 plus 1 times 2 times 1 plus 1 all over 6.

00:22 And it asks to show us this is true.

00:25 Well we know 1 squared is 1 so we need this side to be equal to 1.

00:29 So we have 1 times 2 times 3 over 6 which equals 1.

00:36 This is true this is our base case.

00:43 Now we can write our inductive hypothesis.

00:54 When this is going to be we're going to assume this is true for 1 all the way up top.

01:14 So that's going to be 1 squared plus i squared is equal to i times i plus 1 times 2i plus 1 all over 6.

01:30 So now we can go to our inductive step.

01:33 So for our inductive step we want to consider the i plus 1 case.

01:52 That's going to be 1 squared plus 2 squared plus all the way up to i squared plus i plus 1 squared.

02:04 And this is equal to i times i plus 1 times 2i plus 1 all over 6 plus i plus 1 squared.

02:18 This is by the inductive hypothesis.

02:20 So you see this is equal to this by the inductive hypothesis.

02:30 All right so now our goal is to get an i plus 1 term that looks like this.

02:40 So let's get a common denominator and combine this into one fraction.

02:49 So we see we can do that by having this is equal to i times i plus 1 times 2i plus 1 plus.

03:02 6i plus 1 squared all over 6.

03:10 And we see we have an i plus 1 here.

03:16 We also have an i plus 1 here.

03:18 So we can pull that out as a factor.

03:26 So this is equal to i plus 1 times...

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