How Does Sum of Consecutive Numbers Relate to Cubes of Integers?

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The discussion centers on the relationship between the sum of consecutive integers and the cubes of integers, specifically expressed as n^3 + (n+1)^3. The left-hand side of the equation is represented by the sum of an arithmetic progression, which can be formulated as [(n+1)^2 + 1] + [(n+1)^2 + 2] + ... + (n+1)^2. The user successfully verified their hypothesis through the back of the book but faced challenges in proving it using mathematical induction. Ultimately, they were advised to utilize the formula for the sum of an arithmetic progression to facilitate their proof.

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Homework Statement



Consider the table:

1 = 0 + 1
2 + 3 + 4 = 1 + 8
5 + 6 + 7 + 8 + 9 = 8 + 27
10 + 11 + 12 + 13 + 14 + 15 + 16 = 27 + 64

Guess the general law suggested by these examples, express it in suitable mathematical notation, and prove it.

Homework Equations


[/B]
It's clear that if n is "row" number, then the right-hand side is n^3 + (n+1)^3. The left-hand side is in arithmetic procession, and it can be expressed as [(n+1)^2 + 1] + [(n+1)^2 + 2] + ... + (n+1)^2

The Attempt at a Solution


[/B]
From the equations above, my guess at the general solution is [(n+1)^2 + 1] + [(n+1)^2 + 2] + ... + (n+1)^2=n^3 + (n+1)^3 I verified that this is correct from the back of the book, but I'm having trouble proving it. I've been trying to use induction. The base case holds. Then we want to show that
[((n+1)+1)^2 + 1] + [((n+1)+1)^2 + 2] + ... + (n+1)^2=(n+1)^3 + ((n+1)+1)^3 I'm not sure where to go from here. I tried a few algebraic manipulations, but they didn't seem promising.

SOURCE: Mathematics and Plausible Reasoning by George Polya. Chapter 1, question 2.
 
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EternusVia said:

Homework Statement



Consider the table:

1 = 0 + 1
2 + 3 + 4 = 1 + 8
5 + 6 + 7 + 8 + 9 = 8 + 27
10 + 11 + 12 + 13 + 14 + 15 + 16 = 27 + 64

Guess the general law suggested by these examples, express it in suitable mathematical notation, and prove it.

Homework Equations


[/B]
It's clear that if n is "row" number, then the right-hand side is n^3 + (n+1)^3. The left-hand side is in arithmetic procession, and it can be expressed as [(n+1)^2 + 1] + [(n+1)^2 + 2] + ... + (n+1)^2

The Attempt at a Solution


[/B]
From the equations above, my guess at the general solution is [(n+1)^2 + 1] + [(n+1)^2 + 2] + ... + (n+1)^2=n^3 + (n+1)^3 I verified that this is correct from the back of the book, but I'm having trouble proving it. I've been trying to use induction. The base case holds. Then we want to show that
[((n+1)+1)^2 + 1] + [((n+1)+1)^2 + 2] + ... + (n+1)^2=(n+1)^3 + ((n+1)+1)^3 I'm not sure where to go from here. I tried a few algebraic manipulations, but they didn't seem promising.

Indeed, induction is often used to proof things like this. However, you might want to use the following formula: https://en.wikipedia.org/wiki/Arithmetic_progression#Sum
 
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Math_QED said:
Indeed, induction is often used to proof things like this. However, you might want to use the following formula: https://en.wikipedia.org/wiki/Arithmetic_progression#Sum

That worked! I would post the process, but it follows almost immediately from applying the formula for the sum of an arithmetic progression.

Thank you
 
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