ASAP: Identifying the series (1 Viewer)

[Run hard@thehsc]

ASAP: I have to identify a series for the number of handshakes that can be made for 'n' people. Is the below series right?
0 + …. + (n-2) + (n-1). I have had some people tell me that it is 0 + 1 + ... + (n-1), but I believe that a group with one person can make zero handshakes - so what would the best way to represent this series be? Clarification would be awesome.


Also, what is the relation between the inductive step and the initial statement in mathematical induction?

 

[cossine]

I think you are looking at metcalfe rule or law.

initial statement is the first statement

e.g. we prove true for the n = 1

inductive statement is assume true for n = k .

 

[Run hard@thehsc]

@cossine oh ok. For the series, would it be represented as 0 + .... + (n-2) + (n-1)?

 

[cossine]

ASAP: I have to identify a series for the number of handshakes that can be made for 'n' people. Is the below series right?
0 + …. + (n-2) + (n-1). I have had some people tell me that it is 0 + 1 + ... + (n-1), but I believe that a group with one person can make zero handshakes - so what would the best way to represent this series be? Clarification would be awesome.


Also, what is the relation between the inductive step and the initial statement in mathematical induction?

To be honest you have lost me with your question.

So, 0 + …. + (n-2) + (n-1) and 0 + 1 + ... + (n-1) are the same.

I believe that a group with one person can make zero handshakes. Of course it will be zero.

It seems like the formula for is based on sum of an arithmetic sequence. (a +l)/2 * n

In this case formula is n(n-1)/2 as per wikipedia.

 

[CM_Tutor]

Suppose we have a group of 4 people, conveniently named A, B, C, and D.

If everyone shakes hands with everyone else, the number of handshakes undertaken by A is 3 (A with B, A with C, and A with D)... but the total number of handshakes is 6 (A with B, A with C, A with D, B with C, B with D, and C with D).

If you are looking at handshakes for A, the number will increase by 1 with every new person added to the group (A has hanshakes with everyone from the previous case plus one new handshake).

If you are talking total handshakes, however, then I get the sequence to be:

1 person ===> 0 handshakes
2 people ===> 1 handshake
3 people ===> 3 handshakes
4 people ===> 6 handshakes
5 people ===> 10 handshakes
6 people ===> 15 handshakes
7 people ===> 21 handshakes

The pattern shows an increase by n - 1 handshakes so that h_n, the handshakes amongst n people is given by h_n = h_n-1 + n - 1 with h₁ = 0. Since this number is also a binomial coefficient, the solution will be:

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