by mathematical induction, to prove even number of persons is odd number of shakehands?

Answer 1

If you have two people, they shake hands once (odd).
If you add two more people, they will each have an even number of handshakes with the existing people, plus one more handshake for themselves. odd + even + odd = even. So they will change the parity from odd to even.

So I've proven it false by contradiction.

Repeat the process and you will add an even number of handshakes to the existing participants, plus 1 more handshake for the pair...again changing the parity from even to odd.

2 people = 1 handshake (odd)
4 people = 6 handshakes (even)
6 people = 15 handshakes (odd)
8 people = 28 handshakes (even)
10 people = 45 handshakes (odd)
12 people = 66 handshakes (even)
etc.

The general formula is n choose 2, or n(n-1) / 2.

For the case of:
n = 2 --> 2 * 1 / 2 = 1
n = 4 --> 4 * 3 / 2 = 6
n = 6 --> 6 * 5 / 2 = 15
n = 8 --> 8 * 7 / 2 = 28
n = 10 --> 10 * 9 / 2 = 45
n = 12 --> 12 * 11 / 2 = 66
etc.

Are you sure you stated the problem correctly? Perhaps you meant for groups of 2, 6, 10, etc. adding four people?

Edit:
Okay, I've reread your question and the only interpretation that might work is if you are counting the handshakes for *one* person. For example, if I am one of two people, then I shake hands with one person. Now if you add two more people, I shake hands with three people. If you add two more people, I shake hands with five people, etc.

First you prove that with 2 people, I shake hands with one person (2-1). This is pretty obvious. Now assume you have n people and an odd number of handshakes. This can be represented as 2k + 1 handshakes, where k is an integer. You need to prove that with n+2 people, you still have an odd number of handshakes. Since you are adding 2 new handshakes, this would be 2k + 3 handshakes which is also 2(k+1) + 1. If k is an integer, so will k + 1. As a result 2(k+1) + 1 will also be an odd number, so by mathematical induction you have proven that in a room with an even number of people, you will shake hands an odd number of times.

Answer 2

Nope.
A shakes hands with B, C & D=3
B shakes hands with C & D=2
C shakes hands with D=1
3+2+1=6 even

If n is even BUT NOT DIVISIBLE BY 4 the # of handshakes will be odd.
For n=6, there will be 5+4+3+2+1=15 handshakes.

Answer 3

N = n(n+1)/2
Odd "shakehands" occur when n = {2,5,6,9,10,13,14,...)

As is readily obvious not all of these numbers are even, and even numbers are not all included.

Answer 4

no its still even