Prove that 1^2+3^2+5^2+⋯+(2 n+1)^2=(n+1) (2 n+1)(2 n+3) / 3...

Key Concepts

Mathematical Induction

Mathematical induction is a proof technique used to establish the truth of an infinite number of cases by first verifying a base case and then proving that if the statement holds for an arbitrary case, it holds for the next case. In this context, it is used to show that the formula for the sum of the squares of odd numbers is valid for every nonnegative integer n.

Summation of Sequences

Summation of sequences involves finding a closed-form expression that represents the sum of a series of terms. Here, the sequence considered is the squares of odd numbers, and the goal is to derive a formula that expresses the sum in terms of n. Understanding how to manipulate and express sums is essential in topics like series and sequences.

Algebraic Manipulation

Algebraic manipulation refers to the process of rearranging and simplifying expressions to reveal underlying structures or to prove identities. In the problem, this involves expanding the sum in the inductive step, simplifying terms, and ultimately arriving at the closed-form expression. Such skills are fundamental in a wide range of mathematical proofs and computations.

Transcript

00:01 So we're given this statement, we're asked to prove it for when n is greater than zero.

00:05 This is an induction problem, so we want to start with our base case.

00:10 Our base case is going to be n equals zero.

00:17 So if we plug in zero to our formula, two times zero is zero plus one squared, so we have one squared is equal to zero plus one times two times zero plus one times two times zero plus three all over three, which we get is equal to one times one times three, so we get three over three, which equals one.

00:47 So our base case passes.

00:50 Now we go on to our inductive hypothesis.

00:53 So our inductive hypothesis is going to say, well we'll assume that the first i are true, so one squared plus three squared plus five squared all the way up to two times two i plus one squared is equal to i plus one times two i plus one times two i plus three.

01:50 Can't fit it in there.

02:00 All over three.

02:03 So now we go to our inductive step and it says, well let's look at our i plus one.

02:14 So we'll have one squared plus three squared plus five squared plus all the way up to two i plus one squared plus two times two times i plus one, which is going to be two i plus two plus another one squared.

02:46 So we're looking at this case and we know this is equal to i plus one times two i plus one times two i plus three all over three plus two i plus three squared.

03:11 And this is, this whole thing is equal to this by the inductive hypothesis.

03:23 That's how we defined it in our last step...

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