# Need help with Simplification step in Inductive Proof

• leo255
In summary: This is what is called the quadratic equation in two variables:##\frac{k^2(k + 1)^2}{4} + (k + 1)^3 = 0##
leo255

## Homework Statement

Let P (n) be the statement that 1^3 + 2^3 + · · · + n^3 = (n(n + 1)/2)^2 for the positive integer n. Prove inductively.

## The Attempt at a Solution

[/B]
I am skipping a few steps...I just need help here:

1/4K^2(k + 1)^2 + (k + 1)^3

Since I have access to the solution, the next step is this:

1/4(k+1)^2 [K^2 + 4 (k + 1)]

I am confused at how this is gotten to. I appreciate the help.

Thanks.

leo255 said:

## Homework Statement

Let P (n) be the statement that 1^3 + 2^3 + · · · + n^3 = (n(n + 1)/2)^2 for the positive integer n. Prove inductively.

## The Attempt at a Solution

[/B]
I am skipping a few steps...I just need help here:

1/4K^2(k + 1)^2 + (k + 1)^3
Yes, you are skipping some steps.
The first term above is from ##1^3 + 2^3 + \dots + k^3## and the other term above is from adding ##(k + 1)^3## in your induction step.
In the two terms, do you notice that there is a common factor?
leo255 said:
Since I have access to the solution, the next step is this:

1/4(k+1)^2 [K^2 + 4 (k + 1)]

I am confused at how this is gotten to. I appreciate the help.

Thanks.

Ahhh, yes. (k+1).

For some reason, I was trying to factor out k, k^2, etc.

So, from factoring out, I am getting:

1/4(k+1)^2 [k^2 + (k+1)]

I'm still a little confused as to where the 4, in front of the (k + 1) in the brackets is coming from. I assume it's because we are factoring out a 1/4, so we're just multiplying it by 4, to make it equal to one.

leo255 said:
Ahhh, yes. (k+1).

For some reason, I was trying to factor out k, k^2, etc.

So, from factoring out, I am getting:

1/4(k+1)^2 [k^2 + (k+1)]

I'm still a little confused as to where the 4, in front of the (k + 1) in the brackets is coming from. I assume it's because we are factoring out a 1/4, so we're just multiplying it by 4, to make it equal to one.
The part you left out is fouling you up.
The induction hypothesis is:
##1^3 + 2^3 + 3^3 + \dots + k^3 = \frac{k^2(k + 1)^2}{4}##
Now, work from the induction step:
##1^3 + 2^3 + 3^3 + \dots + k^3 + (k + 1)^3 = \frac{k^2(k + 1)^2}{4} + (k + 1)^3##

What happens when you factor out ##(k + 1)^2## from the right side?

## What is an inductive proof?

An inductive proof is a mathematical method used to prove a statement or theorem by showing that it holds for a base case and then showing that if it holds for a certain value, it also holds for the next value. This process is repeated until it can be shown to hold for all values.

## What is the role of simplification in an inductive proof?

Simplification is a crucial step in an inductive proof as it helps to reduce the complexity of the problem at hand. By simplifying the statement or equation, it becomes easier to analyze and prove using the inductive method.

## Why is it important to show all steps of simplification in an inductive proof?

Showing all steps of simplification in an inductive proof helps to ensure that the proof is valid and free of errors. It also allows others to understand and follow the logic of the proof, making it easier to verify its accuracy.

## What are some common techniques for simplifying inductive proofs?

One common technique for simplifying inductive proofs is to use mathematical operations, such as addition, subtraction, multiplication, and division, to manipulate equations. Another technique is to use mathematical properties, such as the distributive property, to simplify expressions. Additionally, simplifying by grouping terms and canceling out common factors can also be useful.

## What are some tips for successfully completing the simplification step in an inductive proof?

Some tips for successfully completing the simplification step in an inductive proof include carefully examining the statement or equation to identify any patterns or relationships, using known mathematical properties and operations to manipulate the equation, and checking for errors or inconsistencies along the way. It is also helpful to break down the problem into smaller, more manageable steps and to double check all calculations and substitutions.

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