Secondary school level mathematical induction

Question

1. 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?

2. 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?

3. 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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

For #3 use the technique of #2: For $n\in \mathbb N \cup \{0\}$ we have $$a^{2n+3}+b^{2n+3}=a^2(a^{2n+1}+b^{2n+1})+(-a^2+b^2)b^{2n+1}=$$ $$=a^2(a^{2n+1}+b^{2n+1})+(a+b)(b-a)b^{2n+1}.$$