Secondary school level mathematical induction 
*

*It is given that 
$$1^3+2^3+3^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}$$


Then, how to find the value of 
$2^3+4^3+\cdots+30^3$? 
Which direction should I aim at?


*Prove by mathematical induction, that $5^n-4^n$ is divisible by 9 for all positive even numbers $n$.


$$5^n-4^n=9m,\text{where $m$ is an integer.}$$
What I am thinking in the $n+1$ step is,
\begin{align}
& 5^{n+2}-4^{n+2} \\
= {} & 5^2(5^n-4^n)+5^24^n-4^{n+2} \\
= {} & 5^2(5^n-4^n)+4^n(5^2-4^2) \\
= {} & 5^29m+4^n9 \\
= {} & 9(5^2m+4^n)
\end{align}
Does this approach make sense?


*Show that $a+b$ is a factor of $a^n+b^n$ where $n$ is a positive odd number.


I am thinking this in the $n+1$ step.
$$a^{2n+1}+b^{2n+1}$$
But then I cannot get it further.
 A: I'll give you a hint for the first one. Since you have asked three different questions, I wait to see if they tell you to split your question in there questions...
1
You already have a formula:
$$1^3 + 2^3 + 3^3 + \ldots  + n^3 = \frac{n^2((n+1)^2)}{4}$$
Hence in your case $n = 30$ but your sum starts from $2^3$ and not from $1^3$, hence you have to subtract $1^3$ (namely one):
$$2^3 + 3^3 + \ldots + 30^3 = \frac{30^2(30+1)^2}{4} - 1 = \frac{30^2\cdot 31^2}{4} - 1 = 216224$$
Edit because I understood nothing
Since you want only the sum of even numbers, then you have:
$$2^3 + 4^3 + 8^3 + \ldots + 30^3 = 2^3(1^3 + 2^3 + 3^3 + \ldots + 15^3) = 115200$$
Thanks to Crostule for notifying me.
A: Hints
1) $2^3+4^3+...+30^3=2^3(1^3+2^3+...+15^3)$ now apply the formula you got.
2) Assume that $9k=5^n-4^n$ then $5^{n+2}-4^{n+2}=25.5^n-16.4^n=25(9k+4^n)-16.4^n=9k'+9.4^n$
3) Change the questions to show that $(a+b)|(a^{2k+1}+b^{2k+1})$ for all $k\in\mathbb{N}$.Let $(a+b)x=a^{2k+1}+b^{2k+1}$ now $a^{2k+3}+b^{2k+3}=a^2.a^{2k+1}+b^2.b^{2k+1}$ again do substitution and it will work out.
A: 1) $2^3 + 4^3 + 6^3 + .... +30^3 = 2^3*1^3 + 2^3*2^3 + 2^3*3^3 + ... + 2^3*15^3 = 2^3(1^3 + 2^3 + .... + 25^3)$
2) Perfect.  You did great.  (Better than I did when I didn't realize it was only true for even numbers.)
3) If $n$ is odd then you don't want $a^{2n+1} + b^{2n+1}$ for the inductive step but either: $a^{n+2} + b^{n+2}$ (much like you did in 2) or $n = 2k+1;$ inductive step on $a^{2(k+1) + 1} + b^{2(k+1)+1}$. 2 1/2) or $n = 2k-1$ and $b^{2k+1} + b^{2k +1}$.  Either way will work.
A: For #3 use the technique of #2: For $n\in \mathbb N \cup \{0\}$ we have $$a^{2n+3}+b^{2n+3}=a^2(a^{2n+1}+b^{2n+1})+(-a^2+b^2)b^{2n+1}=$$ $$=a^2(a^{2n+1}+b^{2n+1})+(a+b)(b-a)b^{2n+1}.$$
