i know this is a kind of big list, but i'm completely lost as to how to do these sort of things and i can't seem to make sense of the book (i've spent about 2 hours trying to figure it out)
any help walking me through these would be greatly appreciated
Prove the following by induction:
A) For all integers n >= 0, the number 5^(2n) - 3n is a multiple of 11
B) Any integer n >= 1, 2^(4n-1) ends with an 8
C) The sum of the cubes of three consecutive positive integers is always a multiple of 9
D) If x >= 2 is a real number and n >= 1 is an integer, then x^n >= nx
E) If n >= 3 is an integer, then 5^n > 4^n + 3^n + 2^n
A) Doesn't work. e.g. n=4?
D)
When x=2 and n=1, Therefore true for n=1.
Assume it's true for n=k.
now for induction, we want to prove that
so here is a standard 'trick' often required for similar questions..
For n=k
Times both sides of the inequality by x...
But for x > 0
Add x to both sides.....
Since
It is implied, therefore, that as required.
Hence if n=k is true, then n=k+1 is true. But since n=1 is true, it implies n=2 is true, which implies n=3 is true...etc. Hence proved by induction.
Well, this is not what you'd call 'elegant', but it's a proof of sorts...
Just use the same 'trick' I did for question D.
E) It's easy to prove that the equation is true for n > 2. So I won't waste time or patronise you by proving that. So assume it is true for n=k. Just deal exclusively with the
multiply each side by 5
But for k > 2
But
(Since )
Therefore it follows that
which implies as required. Hence if n=k is true, then n=k+1 is true. But since n=1 is true, it implies n=2 is true, which implies n=3 is true...etc.
Repeat the process for then . If you want more help, I'd more more than willing to provide it - but can you at least try to say anything you don't understand - or better yet, post some working, that would be great.
Hello, mistykz!
Let the three consecutive cubes be: .C) The sum of the cubes of three consecutive positive integers is always a multiple of 9
We want to show that: . .for some integer
Vertify . . . yes!
Assume .for an integer
Add to both sides:
. .
and we have:
. .
Therefore: . . . . . a multiple of 9
The induction proof is complete.