Prove by Induction: 1^2 + 2^2 + ... + (n-1)^2 < (n^3)/3

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In summary, to prove the given inequality by induction, we first show that A(k+1) is a consequence of A(k) by adding k^2 to both sides of A(k). Then, using the transitive property of <, we show that (k^3)/3 + k^2 < (k+1)^3/3, which proves the inequality. This is because (k+1)^3/3 is greater than k^3/3 + k^2, which is a consequence of A(k). Therefore, the original inequality is also true for n = k+1, which completes the proof by induction.
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Jimmy84
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Homework Statement


Prove by induction

1^2 + 2^2 + ... + (n-1)^2 < (n^3)/3

(This is the problem on page 33 of Apostol's book)



Homework Equations





The Attempt at a Solution



A(k) = 1^2 + 2^2 + ... + (k-1)^2 < (k^3)/3

A(k+1) = 1^2 + 2^2 + ... + k^2 < (k+1)^3/3


Start with A(k) and add k^2 to both sides. This gives the inequality

1^2 + 2^2 + ... + k^2 < (k^3)/3 + k^2


To obtain A(k+1) as a consequence of this, it suffices to show that

k^3/3 + k^2 < (k+1)^3 /3




Now (k+1)^3 /3 = k^3/3 + k^2 +k +1/3 which is greater than k^3/3 + k^2


Which proves the inequality. However I don't understand why does

k^3/3 + k^2 < (k+1)^3 /3 suffice to prove the inequality

Can anyone explain in detail why does that inequality proves it?
 
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  • #2
They're using the transitive property of <. You know

1^2 + 2^2 + ... + k^2 < (k^3)/3 + k^2.

If you can show (k^3)/3 + k^2 < (k+1)^3/3, then you'll have

1^2 + 2^2 + ... + k^2 < (k^3)/3 + k^2 < (k+1)^3/3

Jimmy84 said:

Homework Statement


Prove by induction

1^2 + 2^2 + ... + (n-1)^2 < (n^3)/3

(This is the problem on page 33 of Apostol's book)

Homework Equations


The Attempt at a Solution



A(k) = 1^2 + 2^2 + ... + (k-1)^2 < (k^3)/3

A(k+1) = 1^2 + 2^2 + ... + k^2 < (k+1)^3/3Start with A(k) and add k^2 to both sides. This gives the inequality

1^2 + 2^2 + ... + k^2 < (k^3)/3 + k^2 To obtain A(k+1) as a consequence of this, it suffices to show that

k^3/3 + k^2 < (k+1)^3 /3

Now (k+1)^3 /3 = k^3/3 + k^2 +k +1/3 which is greater than k^3/3 + k^2 Which proves the inequality. However I don't understand why does

k^3/3 + k^2 < (k+1)^3 /3 suffice to prove the inequality

Can anyone explain in detail why does that inequality proves it?
 
  • #3
Thanks a lot, I really appreciate it. :) I have been struggling with this for quite some time.
 

FAQ: Prove by Induction: 1^2 + 2^2 + ... + (n-1)^2 < (n^3)/3

What is the concept of mathematical induction?

Mathematical induction is a method of mathematical proof used to establish that a statement is true for all natural numbers. It involves proving a base case, typically for n = 1, and then showing that if the statement holds for any n, it also holds for n+1.

How does mathematical induction work?

Mathematical induction works by proving a base case, usually for n = 1, and then assuming the statement holds for any n. This assumption, known as the induction hypothesis, is then used to prove that the statement also holds for n+1. This process is repeated until the statement is proven for all natural numbers.

What is the statement being proven by induction here?

The statement being proven by induction in this case is: 1^2 + 2^2 + ... + (n-1)^2 < (n^3)/3.

How do you prove this statement by induction?

To prove this statement by induction, we first show that it holds for the base case n = 1, which is true since 1^2 = 1 < 1^3/3 = 1/3. Then, we assume that the statement holds for any n, and use this assumption to prove that it also holds for n+1. The induction hypothesis can be used to rewrite the statement as (1^2 + 2^2 + ... + (n-1)^2) + n^2 < (n^3)/3 + n^2. By the induction hypothesis, we know that 1^2 + 2^2 + ... + (n-1)^2 < (n^3)/3, so adding n^2 to both sides gives us the desired result.

Why is mathematical induction a valid method of proof?

Mathematical induction is a valid method of proof because it relies on the fundamental principle of mathematical induction, which states that if a statement is true for the first natural number and if it holds for any number n, then it also holds for n+1. This principle is based on the well-ordering property of the natural numbers, which states that every non-empty set of natural numbers has a least element. Therefore, by proving the base case and showing that the statement holds for any n, we can conclude that it holds for all natural numbers.

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