1. Polynomials
Example. The most stupid M.I. in the world, showing that
In polynomial form M.I. is always easy:
so that the proof is complete.
Example 2.
Proof.
Exercise 1. Show that:
i)
ii)
In fact, there's no such a general form for sum of the n-th power of natrual number.
2. Divisibility and integral solution
Example 3.
Let
In more complicated case, some of the grouping method may not work.
Example 4.
Let
Example 5. (CSMO)
It seems to be very fresh to most M2 or even Pure student as you don't know the proposition to be set in order to prove the statement.Think inversely, the next product must be integer if the last one is a multiple, (and of course, an integer) of the next i, then the next term must be integer too.
Therefore the proposition is
Exercise 2. Show that
1)
2) Finish the proof for ex. 5.
Challenge 1. It's given that the first 4 terms are 2,6,20,70,...
a) Give a general formula for the series with proof.
b) The series satisfies
The last example is from HKALE.
Example 6. (HKALE Pure 1998, I5) For
This is a 5-mark question and shouldn't be hard, or even fit perfectly for M2/Pure level, however we need another method of Mathematical Induction:
Theorem. Second principle of M.I. The proposition is true for all natrual n if:
1) n = 1,2,...i is true.
2) If n = j, (j+1), ... i+j is true, then n = i+j+1 is true.
in this case i = 2.
Assume
By relationship between roots and equations, we have
Then
The last term is trivially a positive integer.
Challenge 2. The recurrence relationship of the sequence
Exercise 3. Is it true that for x,y as roots of
Read the rest of this passage:
Part II
Part III
Part IV
No comments:
Post a Comment