Assume that is true. Then look at:
.
Is that clearly a multiple of 3?
I'm trying to prove through induction. (and for 5 in but that's for later)
Done the base case which simplifies to 2
Assume for evaluates to
Inductive prove for evaluates to
But I'm guessing you have to add and minus some value before to incorporate the assumption so I'm guessing we do something here:
Don't know what exactly, pointers?
Sorry about the mundane question. I'm not wired to think the 2 steps ahead for proofs.
I dont understand the stuff you have written....
But if you substituted x = 2 and checked if for the base case, then you are right.(or have taken the easier route )
Induction Hypothesis: Assume the result is true for x = k: This means
We want to prove the Induction Step:
But
By Induction Hypothesis, , 3 clearly divides , thus 3 divides the sum
By the way, you said you were to use induction so that is the way you were answered. But , the product of three consecutive integers (assuming x is an integer). One of them has to be a multiple of 3.
is a little harder. If x is a multiple of 5, then we are done. if x= 5n+1, then x-1= 5n, a multiple of 5, and we are done. If x= 5n+ 2, then neither x-1= 5n+1 nor x+1= 5n+ 3 is a multiple of 5 but is a multiple of 5. If x= 5n+ 3, then neither x-1= 5n+2 nor x+1= 5n+4 is a multiple of 5 but is a multiple of 5. Finally, if x= 5n+4, then x+1= 5n+ 5= 5(n+1) is a multiple of 5.
Hello everyone -
It is perhaps interesting to observe that is divisible by for any prime .
The proof (by induction) follows exactly the same lines as before, noting that, with the exception of the first and last terms, all the coefficients in the expansion of are divisible by , when is prime. This is simply because the numerator of has a prime factor for , but the denominator doesn't.
Grandad