Prove that for all n Îℤ+; $\displaystyle 2^{3n+2} + 5^{n+1}$ is divisible by $\displaystyle 3$.
I've learn how to prove other standard formula but don't really know how tp prove that it is divisible by an integer. Thanks in advance.
Hello,
You can do the basis
Suppose $\displaystyle f(n)=2^{3n+2}+5^{n+1}=3k$, i.e. divisible by 3.
$\displaystyle \implies 2^{3n+2}=3k-5^{n+1}$
$\displaystyle f(n+1)=2^{3(n+1)+2}+5^{(n+1)+1}=2^{3n+3+2}+5^{n+2}$
$\displaystyle f(n+1)=8 \cdot 2^{3n+2}+5^{n+2}=8 \cdot (3k-5^{n+1})+5 \cdot 5^{n+1}$
$\displaystyle f(n+1)=8 \cdot 3k-8 \cdot 5^{n+1}+5 \cdot 5^{n+1}$
$\displaystyle f(n+1)=3 \cdot (8k)+5^{n+1}(-8+5)=3 \cdot (8k)-3 \cdot 5^{n+1}$
Conclude
outline:
Let $\displaystyle P(n): $ "$\displaystyle 2^{3n + 2} + 5^{n + 1}$ is divisible by 3" for all $\displaystyle n \in \mathbb{Z}^+$.
Clearly $\displaystyle P(1)$ is true (Check this!)
Now, assume $\displaystyle P(n) $ is true, that is, $\displaystyle 2^{3n + 2} + 5^{n + 1} = 3k$ for some $\displaystyle k \in \mathbb{Z}$ (which means $\displaystyle P(n)$ is divisible by 3). we need to show that $\displaystyle P(n + 1)$ is divisible by 3.
Now, $\displaystyle P(n + 1):$ $\displaystyle 2^{3(n + 1) + 2} + 5^{(n + 1) + 1} = 8 \cdot 2^{3n + 2} + 5 \cdot 5^{n + 1}$
$\displaystyle = 8 \cdot 2^{3n + 2} \underbrace{+ 8 \cdot 5^{n + 1} - 8 \cdot 5^{n + 1}}_{\text{this is zero}} + 5 \cdot 5^{n + 1}$
$\displaystyle = 8(2^{3n + 2} + 5^{n + 1}) - 3 \cdot 5^{n + 1}$
...
i leave the last step to you
EDIT: Hey Moo! I thought you were going to bed! you couldn't resist stealing another question from me, could you?