$6^n-5n+4$ is divisible by $5 \;$ for all natural numbers $n$.
what I did is:
IA
$A(1):\;6^1-5\cdot1+4=5$ which is true.
IS
$A(n):\; 6^n+5n+4$ is also divisible by $5$.
Show $A(n+1)$ is divisible by $5$
$$A(n+1): \; 6^{n+1}-5(n+1)+4=6^n\cdot 6-5n-5+4$$ After this step I want to get to $A(n)$ but how?
please give tips to solve this problem.
On
Another alternate method:
The right-most digit of $6^n$ is $6$
Adding $4$ gives a right-most digit of $0$; this new number is divisible by $5$
Subtracting $5n$, a number divisible by $5$, produces a number divisible by $5$.
On
Another way is to look at the difference between consecutive terms (this is what is needed for an induction proof):
$(6^{n+1}-5(n+1)+4) -(6^n-5n+4) =(6^{n+1}-6^n)-5 =6^n(6-1)+5 =5(6^n-1) $ so the difference between consecutive terms is always divisible by $5$.
Since the first term is divisible by $5$, all the terms are.
You have $A(n+1) = 6\cdot 6^n - 5n-5+4 = 5\cdot 6^n + 6^n - 5n - 5 + 4$. But $5\cdot 6^n$ and $-5$ are both divisible by $5$, so $A(n+1)$ is divisible by $5$ if and only if the other terms are; those terms are $6^n - 5n + 4 = A(n)$.
An alternate way of approaching this problem is directly, using the binomial theorem: $$6^n - 5n + 4 = (5+1)^n - 5n + 4.$$ Every term in the expansion of $(5+1)^n$ except for the final term is divisible by $5$, and $-5n$ is divisible by $5$. So the leftover terms are $1+4=5$, which is also divisible by $5$.