Proof by induction

[Randyyy]

Hey, Mathhelp! Is my proof valid? I am new to induction so I am not too confident what I have done is correct.


Prove that 3ⁿ-1 is divisible by 2 [MATH]\forall n \in\mathbb{Z}^+ ,n\geq 1[/MATH]S(n)=3ⁿ-1=2k, [MATH]k\in\mathbb{Z}[/MATH]Base-case n=1: 3¹-1=2*1, true for n=1
S(p)=3^p-1=2k
Assuming it is true for n=p, then it is true for n=p+1.
S(p+1)= 3^p+1-1=2r, [MATH]r\in\mathbb{Z}[/MATH]S(p)=3(3^p-1)=3*2k
S(p)=3^p+1-3=6k
S(p)=3^p+1-1=6k+2
S(p)=3^p+1-1=2(3k+1)
(3k+1)=r and hence:
S(p)=3^p+1-1=2r

 

[pka]

Prove that 3ⁿ-1 is divisible by 2 [MATH]\forall n \in\mathbb{Z}^+ ,n\geq 1[/MATH]

Lets's suppose that \(3^p-1=2k\) where \(\{p,k\}\subset\mathbb{Z}^+\)
Look at
\( \begin{align*}3^{p+1}-1&=3\cdot 3^p-1 \\&=3\cdot 3^p-3+2\\&=3\cdot (3^p-1)+2\\&=3\cdot (2k)+2\\&=2(3k+1)\text{ which is even} \end{align*}\)

 

[Randyyy]

Lets's suppose that \(3^p-1=2k\) where \(\{p,k\}\subset\mathbb{Z}^+\)
Look at
\( \begin{align*}3^{p+1}-1&=3\cdot 3^p-1 \\&=3\cdot 3^p-3+2\\&=3\cdot (3^p-1)+2\\&=3\cdot (2k)+2\\&=2(3k+1)\text{ which is even} \end{align*}\)

Thank you