Answer:
Step-by-step explanation:
First, observe that:
[tex](n+1)^2(2(n+1)^2-1)=(n^2+2n+1)(2n^2+4n+1)=2n^4+8n^3+11n^2+6n+1[/tex]
We will prove by mathematical induction that, for every natural,
[tex]1^3+3^3+5^3+......(2n-1)^3=n^2(2n^2-1)[/tex]
We will prove our base case (when n=1) to be true.
Base case:
[tex]1^3+3^3+5^3+......(2n-1)^3=1=n^2(2n^2-1)[/tex]
Inductive hypothesis:
Given a natural n,
[tex]1^3+3^3+5^3+......(2n-1)^3=n^2(2n^2-1)[/tex]
Now, we will assume the induction hypothesis and then use this assumption, involving n, to prove the statement for n + 1.
Inductive step:
[tex]1^3+3^3+5^3+......(2(n+1)-1)^3=1^3+3^3+5^3+......(2n+1)^3=\\=n^2(2n^2-1)+(2n+1)^3=2n^4-n^2+8n^3+12n^2+6n+1=2n^4+8n^3+11n^2+6n+1[/tex]
Then, by the observation made at the beginning of this proof, we have that
[tex]1^3+3^3+5^3+......(2(n+1)-1)^3=(n+1)^2(2(n+1)^2-1[/tex]
With this we have proved our statement to be true for n+1.
In conlusion, for every natural n
,
[tex]1^3+3^3+5^3+......(2n-1)^3=n^2(2n^2-1)[/tex]
b)
Observe that [tex]3\mid (n^3-n+3) \iff 3\mid (n^3-n) \iff 3\mid n(n^2-1)[/tex]
Then,
Therefore, for every [tex]n\in \mathbb{N}), 3\mid (n^3-n+3)[/tex]
c)
We will prove by mathematical induction that, for every natural n>3,
[tex](n-1)^2<2n^2.[tex]
We will prove our base case (when n=4) to be true.
Base case:
[tex](n-1)^2=(4-1)^2=9<32=2*4^2=2n^2 [/tex]
Inductive hypothesis:
Given a natural n>4,
[tex](n-1)^2<2n^2.[tex]
Now, we will assume the induction hypothesis and then use this assumption, involving n, to prove the statement for n + 1.
Inductive step:
[tex]((n+1)-1)^2=((n-1)+1)^2=(n-1)^2+2(n-1)+1<2n^2+2(n-1)+1=2n^2+2n- 1<2n^2+2n+1 =2(n+1)^2.[tex]
With this we have proved our statement to be true for n+1.
In conclusion, for every natural n>3,
[tex](n-1)^2<2n^2.[tex]
