# Inductive Proof on Well Known Sum

• Shackman
In summary: So in summary, you have not done a proof like this before, and you are looking for a nudge in the right direction.
Shackman

## Homework Statement

Use induction to prove the equation
(1+2+...+n)^2 = 1^3 + 2^3 + ... + n^3

2. The attempt at a solution
I've done three inductive proofs previous to this one where I showed that the equation was true for some case (usually n=1), then assumed it was true for n, and proved it was true for n+1. This proof doesn't seem to fit that form though because for the other proofs I could write it as the right hand side of the equation (since it's assumed that it is true for n) plus the right hand side of the equation when n+1 is plugged in. But in this case..

(1+2+...+n+(n+1))^2 != (1+2+...+n)^2 + (n+1)^2

So, I guess I haven't done a proof like this and could use a nudge in the right direction..

What is $$\sum^{n}_{i=1} i$$?

n(n+1)/2

In an effort to see where you're going with this, I thought I'd do the same for i^2 and got

n^2(n^2 + 1)/4

Either this isn't correct or I'm not sure where you're headed.

Last edited:
Shackman said:
n(n+1)/2

In an effort to see where you're going with this, I thought I'd do the same for i^2 and got

n^2(n^2 + 1)/4

Either this is correct or I'm not sure where you're headed.

Well, the second part isn't correct at all ($\sum^{n}_{i=1} i^2 = \frac{n(n+1)(2n+1)}{6}$), but that wasn't really my point. Look at the left hand side of your proposed equality.

Alright, so the left hand side is equal to ((n(n+1))/2)^2 or (n^4 +2n^3 +n^2)/4...

So prove that (n(n+1)/2)^2 is equal to the right hand side. Much easier to verify.

That is a proof of the equality but it isn't inductive though. Inductive proofs always follow the form:
1) prove the base case
2) write the induction hypothesis for the case of n (assume its true basically)
3) use 2) to prove for n+1

Shackman said:
(1+2+...+n+(n+1))^2 != (1+2+...+n)^2 + (n+1)^2
As you say, that is not the correct formula for the square of two things. Fortunately, you know the right formula for the square of two things... (right?)

You're talking about the binomial theorem right?

## 1. What is inductive proof on well known sum?

Inductive proof on well known sum is a mathematical proof technique used to demonstrate that a statement is true for all natural numbers. It involves showing that the statement is true for a base case (usually n=1) and then using the assumption that it is true for some arbitrary value n to prove that it is also true for n+1.

## 2. How is inductive proof on well known sum different from other proof techniques?

Inductive proof on well known sum is different from other proof techniques, such as direct proof or proof by contradiction, because it relies on the assumption that the statement is true for some arbitrary value rather than directly proving it for all values.

## 3. Can any statement be proven using inductive proof on well known sum?

No, not every statement can be proven using inductive proof on well known sum. This technique is only applicable to statements that can be expressed as a summation of natural numbers.

## 4. Why is it important to use a base case in inductive proof on well known sum?

The base case is important in inductive proof on well known sum because it provides a starting point and allows us to establish the truth of the statement for the first value of n. Without a base case, we cannot use the inductive step to prove the statement for all values of n.

## 5. Are there any limitations to inductive proof on well known sum?

Yes, there are some limitations to inductive proof on well known sum. This technique can only be used to prove statements that involve natural numbers and may not be applicable to more complex mathematical concepts. It also requires the statement to be well-known and easily understood, which may not always be the case.

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