# Proof by mathematical induction.

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• June 23rd 2011, 08:08 PM
Pupil
Proof by mathematical induction.
Show that the statement holds for all positive integers n:
$1^{2} + 2^{2}+...(n-1)^{3} < \frac{1}{4}n^{4} < 1^{3} +2^{3}+...n^{3}$

Firstly, I'll start with $n = 1$
$(1-1)^{3} < \frac{1}{4}(1)^{4} < (1)^{3}$
$0 < \frac{1}{4} < 1$
Therefore, 1 ϵ S.
Assuming k ϵ S:
$1^{2} + 2^{2}+...(k-1)^{3} < \frac{1}{4}k^{4} < 1^{3} +2^{3}+...k^{3}$
But, k ϵ S implies that k + 1 ϵ S, therefore:
$1^{2} + 2^{2}+...(k-1)^{3} +((k+1)-1)^{3} < \frac{1}{4}(k+1)^{4} < 1^{3} +2^{3}+...k^{3} + (k+1)^{3}$
Simplify:
$2k^{3} - 3k^{2} + 3k - 1 < \frac{1}{4}k^{4} + k^{3} + \frac{3}{2}k^{2} + k + 1 < 2k^{3} + 3k^{2} + 3k + 1$

Therefore, k + 1 ϵ S and the statement holds for all positive integers n and all positive integers are in S.

Have I proven this correctly by using mathematical induction? I haven't dealt with an equality yet in all the problem sets that I've done before. Thanks in advance.
• June 23rd 2011, 08:39 PM
godelproof
Re: Proof by mathematical induction.
$1^{2} + 2^{2}+...(k-1)^{3} +((k+1)-1)^{3} < \frac{1}{4}(k+1)^{4} < 1^{3} +2^{3}+...k^{3} + (k+1)^{3}$ is what you are trying to prove by induction, NOT something you already know to be true! You want to prove $1^{2} + 2^{2}+...(k-1)^{3} < \frac{1}{4}k^{4} < 1^{3} +2^{3}+...k^{3}$ $\Longrightarrow$ $1^{2} + 2^{2}+...(k-1)^{3} +((k+1)-1)^{3} < \frac{1}{4}(k+1)^{4} < 1^{3} +2^{3}+...k^{3} + (k+1)^{3}$. How does it go from there?
• June 23rd 2011, 09:14 PM
Pupil
Re: Proof by mathematical induction.
Quote:

Originally Posted by godelproof
$1^{2} + 2^{2}+...(k-1)^{3} +((k+1)-1)^{3} < \frac{1}{4}(k+1)^{4} < 1^{3} +2^{3}+...k^{3} + (k+1)^{3}$ is what you are trying to prove by induction, NOT something you already know to be true! You want to prove $1^{2} + 2^{2}+...(k-1)^{3} < \frac{1}{4}k^{4} < 1^{3} +2^{3}+...k^{3}$ $\Longrightarrow$ $1^{2} + 2^{2}+...(k-1)^{3} +((k+1)-1)^{3} < \frac{1}{4}(k+1)^{4} < 1^{3} +2^{3}+...k^{3} + (k+1)^{3}$. How does it go from there?

That is where I seem to mess up. I haven't dealt with inequalities in my problem sets.

So, I'll assume that:
$1^{2} + 2^{2}+...(k-1)^{3} = \frac{1}{4}(k+1)^{4}$ and $1^{3} +2^{3}+...k^{3} = (k-1)^{3}$. Showing that is not true would prove validity of the original statement, correct?
• June 23rd 2011, 09:25 PM
BAdhi
Re: Proof by mathematical induction.
What I don't understand his how you explain the statement $1^2+2^2+...(n-1)^3$. I mean from where does the power change from 2 to 3. Also I think that the statement should be written as $1^2+2^2+...+(n-1)^3$ not $1^2+2^2+...(n-1)^3$.
• June 23rd 2011, 10:12 PM
Pupil
Re: Proof by mathematical induction.
Quote:

Originally Posted by BAdhi
What I don't understand his how you explain the statement $1^2+2^2+...(n-1)^3$. I mean from where does the power change from 2 to 3.

That is how it asks in my book.

Quote:

Originally Posted by BAdhi
Also I think that the statement should be written as $1^2+2^2+...+(n-1)^3$ not $1^2+2^2+...(n-1)^3$.

Yes, you're correct with the notation. I made a typo.