Prove that $\forall n \in \mathbb{N}: 4^n + 6n - 10$ is divisible by $18$ 
Prove that $\forall n \in \mathbb{N}: 4^n + 6n - 10$ is divisible by $18$

Base case: for $n = 1: 4^1 +6\cdot 1 - 10 = 0$ is divisible by 18.
Inductive Assumption: Assume that for for some $k \in \mathbb{N} :4^k +6k-10$
Proving that $4^{k+1}+6(k+1)-10$ is divisible by 18
$4^{k+1}+6(k+1)-10= 4^k \cdot 4 + 6k + 6 - 10$ $$= \color{green}{4^k +6k-10} + 3
 \cdot 4^k + 6$$ 
The first term is divisible by 18 according to the inductive assumption and I have to find a way to manipulate the second term to be divisible by 18.
There is a similar question posted, but the answers for it are substituting another expression $18m$ instead of manipulating $4^{k+1}+6(k+1)-10$. I want to know how to solve this without substituting 
 A: HINT
If $4^k + 6k -10$ is divisible by $18$, then $4(4^k + 6k -10)$ is divisible by $18$ as well.
Subtract the resulting expression from the expression you get when plugging in $k+1$:
$$4^{k+1} +6(k+1)-10 - 4(4^k + 6k -10) = $$
$$4 \cdot 4^k +6k+6-10 - 4 \cdot 4^k -24k + 40 = $$
$$-18k+36$$
Since $-18k+36$ is clearly divisible by $18$, and since $4(4^k + 6k -10)$ is divisible by $18$, it follows that $4^{k+1} + 6(k+1) -10$ is divisible by $18$ as well.
A: let $$T(n)=4^n+6n-10$$ then $$T(n+1)=4^{n+1}+6(n+1)-10$$ and we get
$$T(n+1)-T(n)=3(4^n+2)$$ thus
$$T(n+1)=T(n)+3(4^n+2)$$
it is clear then $3|4^n+2$ so $9|3(4^n+2)$ and $4^n+2$ is even (note that $$18|T(n)$$ via induction
A: One approach is to observe that
$$3\cdot 4^k+6 = 6(2^{2k-1}+1).$$
Now, it is easy to prove by induction that $2^{2k-1}+1$ is a multiple of $3$ ( note that $2^{2k+1}+1 = 4(2^{2k-1}+1)-3$).
A: \begin{align}3\cdot 4^k+6 &= 6(2\cdot4^{k-1}+1) \\
&=6(3\cdot4^{k-1}+1-4^{k-1})\\
&=6(3\cdot4^{k-1}+(1-2^{k-1})(1+2^{k-1}))\\
&=6(3\cdot4^{k-1}-(2^{k-1}-1)(1+2^{k-1}))\end{align}
Notice that $2^{k-1}$ is not divisible by $3$, hence either $(2^{k-1}-1)$ is divisible by $3$ or $(2^{k-1}+1)$ is divisible by $3$. 
Hence $(3\cdot4^{k-1}-(2^{k-1}-1)(1+2^{k-1}))$ is divisble by $3$ and $6(3\cdot4^{k-1}-(2^{k-1}-1)(1+2^{k-1}))$ is divisible by $18$.
Remark: Using modulo arithmetic makes the task much easier.
A: If $4^k +6k-10$ is divisible by $18$ (inductive hypothesis)
then we can write $4^k +6k-10=18t$ for some integer $t$. That is
$4^k=18t-6k+10$ and substituting in what we have to proof we get
$4^{k+1}+6(k+1)-10=4\cdot 4^{k}+6k-4=$
$=4(18t-6k+10)+6k-4=72t-24k+40+6k-4=18(4t-k+2)$
so also $4^{k+1}+6(k+1)-10$ is a multiple of $18$
Hope it is useful
A: For $n=1$ we have $4^1+6(1)-10=0$, which is divisible by $18$. If there exists $k \in \mathbb{N}$ such that $4^n+6n-10=18k$, for some $n \in \mathbb{N}$, then $$4^{n+1}+6(n+1)-10=4\cdot 4^n+6n+6-10=(4\cdot 4^n+24n-40)-24n+40+6n-4=4\cdot 18k-18n+36=18(k-n+2),$$ and since $m-n+2 \in \mathbb{Z}$ you have that $4^{n+1}+6(n+1)-10$ is divisible by 18. 
By induction, the result follows. 
A: Note that $$3\cdot 4^k+6=3\cdot(4^k+2)$$
Now modulo $3$ we have $$4^k+2\equiv 1^k-1\equiv 0 \bmod 3$$i.e. $4^k+2$ is divisible by $3$
