The Handshaking question

There are 4 people. If first person shakes with 3 others, and second person shakes with 2 others, and third person shakes with last person. Total handshakes = 3 + 2 + 1 = 6

1. Suppose there were 100 people.

2. Suppose there were "n" people.

PLEASE EXPLAIN

There are 4 people.
1st person shakes with rest the 3.
2nd person shakes with the other 2 [since he already shook with the 1st person]
3nd person shakes with the remaining 1 [Since he already shook with 1st and 2nd person]
4th person shakes with 0 other [since he already shook with 1st 2nd and 3rd person]

So total shakes = 0 +1 +2 + 3 = 6 [i rearranged it from lesser to greater]
If you observe, u see that the numbers increase from 0 to n-1. where n is the total number of people.
So basically you have to add up form 0 to n-1 going up 1 in each step.

Easier way to do that is using arithmetic series.
this is an arithmetic series where the first term is 0 and last term is (n-1). Common difference is (3-2)=(1-0) = 1.
You can put it into the formula and find the sum.

Sum of an arithmatic series,S = (n/2){2a + (n-1)d}
a = first term (in this case it's zero)
n = total number of term (in this case number of people)
d = common difference (in this case it's 1)

So,
1) for 100 people,
S =(n/2) {2a + (n-1)d} = (100/2) {0 + (100-1)1} = 50 * 99 = 4950
Answer: 4950

2) For n people
S =(n/2) {2a + (n-1)d} = S =(n/2) {0 + (n-1)1} = (n/2) {n-1} = {n(n-1)}/2
Answer: {n(n-1)}/2

It's saying every person is shaking hands with every other person. There is a pattern which I just saw intact to answer if easily. With the four person question you added every number below four. Do the same with 100 people. I would do the math, but doing the math in my head I got 5050 handshakes. For n people, idk, I guess you could say the sum of every positive number that is less than the number of people.