I need to show 1^3 + 2^3 + ... + n^3 < (1/2)n^4 for all n in N and n >=3.

I want to use mathematical induction, but I don't know if I need to use the first Mathematical induction or the second one?

Thanks

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- Feb 3rd 2010, 06:16 PMinthequestofproofsmathematical induction for an inequality
I need to show 1^3 + 2^3 + ... + n^3 < (1/2)n^4 for all n in N and n >=3.

I want to use mathematical induction, but I don't know if I need to use the first Mathematical induction or the second one?

Thanks - Feb 3rd 2010, 06:34 PMArchie Meade
__If__$\displaystyle 1^3+2^3+....+n^3 < \frac{1}{2}n^4,\ n\ge3$

then hopefully $\displaystyle 1^3+2^3+....+n^3+(n+1)^3 < \frac{1}{2}(n+1)^4$

Proof

$\displaystyle (n+1)^3=n^3+3n^2+3n+1$

$\displaystyle (n+1)^4=n^4+4n^3+6n^2+4n+1$

Then

$\displaystyle \frac{1}{2}n^4+(n+1)^3=\frac{1}{2}n^4+n^3+3n^2+3n+ 1$

which is less than $\displaystyle (n+1)^4$

Hence,__if__$\displaystyle 1^3+2^3+...+n^3 < \frac{1}{2}n^4$

__then__$\displaystyle 1^3+2^3+....+(n+1)^3$ is__certainly__< $\displaystyle \frac{1}{2}(n+1)^4$

The inductive process is validated for the hypothesis.

True for n=3 ?

$\displaystyle 1^3+2^3+3^3=36$

$\displaystyle \frac{1}{2}3^4=40.5$ Proven - Feb 4th 2010, 08:45 AMnovice
We are to prove that $\displaystyle 1+2^3+3^3+...+n^3 < \frac{1}{2}n^4$, for $\displaystyle n\geq 3$

The sum of cube series $\displaystyle 1+2^3+3^3+...+n^3$ is equal to $\displaystyle [\frac{n(n+1)}{2}]^2$

We can restate the question as follows:

$\displaystyle [\frac{n(n+1)}{2}]^2< \frac{1}{2}n^4$, for $\displaystyle n\geq 3$

Proof:

Base case: For integer $\displaystyle k=3$,$\displaystyle [\frac{k(k+1)}{2}]^2=36< \frac{1}{2}k^4=\frac{81}{2}$

Induction hypothesis: Suppose for every integer $\displaystyle k>3, k\in \mathbb{N}, [\frac{k(k+1)}{2}]^2< \frac{1}{2}k^4$

Then

$\displaystyle (k+1)^3+[\frac{k(k+1)}{2}]^2< \frac{1}{2}(k+1)^4$

LHS: $\displaystyle (k+1)^3+[\frac{k(k+1)}{2}]^2$

=$\displaystyle k^3+3k^2+3k+1+\frac{k^2(k+1)^2}{4}$

=$\displaystyle k^3+3k^2+3k+1+\frac{k^2(k^2+2k+1)}{4}$

=$\displaystyle k^3+3k^2+3k+1+\frac{k^4+2k^3+k^2}{4}$

=$\displaystyle \frac{k^4}{4}+\frac{3k^3}{2}+\frac{13k^2}{4}+3k+1$

RHS: $\displaystyle \frac{1}{2}(k+1)^4$

=$\displaystyle \frac{k^4}{2}+2k^3+3k^2+2k+\frac{1}{2}$

Putting LHS and RHS together:

$\displaystyle \frac{k^4}{4}+\frac{3k^3}{2}+\frac{13k^2}{4}+3k+1< \frac{k^4}{2}+2k^3+3k^2+2k+\frac{1}{2}$

$\displaystyle \frac{k^2}{4}+k+\frac{1}{2}<\frac{k^4}{4}+\frac{k^ 3}{2}$

Hence, by induction hypothesis, $\displaystyle 1+2^3+3^3+...+n^3 < \frac{1}{2}n^4$, for all $\displaystyle n\geq 3$ in $\displaystyle \mathbb{N}$ - Feb 4th 2010, 11:17 AMinthequestofproofsquestion
After you put the LHS and RHS together, how do you simplify the inequality?\

thanks for your detailed work - Feb 4th 2010, 12:34 PMnovice
Simple algebra. Move the 1st and 2nd terms to the right hand side, and move the last term from the right hand side to the left, etc.

- Feb 4th 2010, 12:39 PMinthequestofproofsquestion about inequality
THanks!! it makes sense. My new question is that if what we are trying to prove the < (inequality), it seems that the RHS < LHS, because some terms are larger than others. However, there are other terms that are smaller than others. So, even though it seems logical that the inequality is true, is it enough to state it like that, or is there a need of more explanation to justify the inequality?

- Feb 4th 2010, 02:09 PMnovice
- Feb 4th 2010, 02:28 PMArchie Meade
If all the steps to here are understandable, inthequestforproofs,

then you can say $\displaystyle \frac{k^4}{4} > \frac{k^2}{4}$

definately $\displaystyle k^4 > k^2$ for k >1

Also, we can ask... is $\displaystyle \frac{k^3}{2} > k+0.5$ ?

So, is $\displaystyle \frac{k^3}{2} > \frac{2k+1}{2}$ ?

Is $\displaystyle k^3 > 2k+1$ ?

Is $\displaystyle k(k^2) > 2(k+0.5)$ ?

if k > 2, then $\displaystyle k^2 > k+0.5$