I have a question:
Prove that is divisible by 10 for all .
Since my current chapter is on induction, I figure it has something to do with induction, I don't really know. But all the induction I've learned in this chapter looks something like
someone may offer a shorter solution, but here's the way I did it ...
... true for
since the statement is true for , then can be written in the form , where is an integral value ...
the first term is a multiple of 10, as is the second term.
for the 3rd term, ...
if is even, then is a multiple of 10.
if is odd, then is even, making the 3rd term also a multiple of 10.
Yes, skeeter's method is exactly how I did it also.
I would word it slightly differently, but that is just my way!
is divisible by 10 ?
being divisible by 10 causes to also be divisible by 10 ?
If we can show that must be divisible by 10 if is,
then we've shown that if P(k) is true for k=2, this causes P(k) to be true for k=3,
which causes P(k) to be true for k=4....
a never-ending chain of cause and effect.
Then P(k) being true for k=2, causes P(k) to be true for all k above 2
in the natural number system.
We never know that the original statement is true until after we have
validated P(k+1) and checked the formula for k=2.
K and k+1 are used instead in the notation instead of n and n+1 merely to focus in on
all pairs of adjacent terms.
We can also get to the result without induction.
The RHS contains the product of three consecutive integers, so one of them must be even, so is divisible by 2.
If one of these three is divisible by 5 also then we are done, so assume that this isn't the case, in which case either is 1 less than a multiple of 5 or is 1 more than a multiple of 5.
Taking the first of these possibilities,
let say, then
so which is divisible by 5, and in which case so is and so is divisible by 10. Similarly for the other possibility.