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
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
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 $\displaystyle \mathbb{N}$
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?
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 $\displaystyle k^2 > k+0.5$