Question

Use mathematical induction to prove that the formula

1^3+3^3+5^3+......(2n-1)^3=n^2(2n^2-1) is true for all natural n.
(b) show tha n^3-n+3 is divisible by 3 for all natural numbers n.
(c) usw mathematical induction to prove that (n-1)^2<2n^2 for all natural numbwrs n>3

Answer 1

Explanation:

First, observe that:

`(n+1)^2(2(n+1)^2-1)=(n^2+2n+1)(2n^2+4n+1)=2n^4+8n^3+11n^2+6n+1`

We will prove by mathematical induction that, for every natural,

`1^3+3^3+5^3+......(2n-1)^3=n^2(2n^2-1)`

We will prove our base case (when n=1) to be true.

Base case:

`1^3+3^3+5^3+......(2n-1)^3=1=n^2(2n^2-1)`

Inductive hypothesis:

Given a natural n,

`1^3+3^3+5^3+......(2n-1)^3=n^2(2n^2-1)`

Now, we will assume the induction hypothesis and then use this assumption, involving n, to prove the statement for n + 1.

Inductive step:

`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`

Then, by the observation made at the beginning of this proof, we have that

`1^3+3^3+5^3+......(2(n+1)-1)^3=(n+1)^2(2(n+1)^2-1`

With this we have proved our statement to be true for n+1.

In conlusion, for every natural n

,

`1^3+3^3+5^3+......(2n-1)^3=n^2(2n^2-1)`

b)

Observe that
`3\mid (n^3-n+3) \iff 3\mid (n^3-n) \iff 3\mid n(n^2-1)`

Then,

1. If
   `n\equiv0\mod3 \implies 3\mid n(n^2-1)`
2. If
   `n\equiv1\mod3 \implies n^2\equiv1\mod3 \implies (n^2-1)\equiv 0 \mod3 \implies 3\mid n(n^2-1)`
3. If
   `n\equiv-1\mod3 \implies n^2\equiv1\mod3 \implies (n^2-1)\equiv 0 \mod3 \implies 3\mid n(n^2-1)`

Therefore, for every
`n\in \mathbb{N}), 3\mid (n^3-n+3)`

c)

We will prove by mathematical induction that, for every natural n>3,

`(n-1)^2<2n^2.[tex] </p><p>We will prove our base case (when n=4) to be true. </p><p><strong>Base case: </strong></p><p>[tex](n-1)^2=(4-1)^2=9<32=2*4^2=2n^2`

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]