Easy maths question

Thanks, I'm looking for a formula that gives the answer straight off.

Just guessing it seems to be N choose 2, that googles to the right answer for 100

Is that right?

In excel you can use below formula assuming  your data starts from A1:

=SUMPRODUCT(LARGE(A1:A100,ROW(INDIRECT("2:100"))))

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I assume we don't need arrays.

If N choose 2 is the right answer then i can just use that formula, i assume Excel has the function, else a factorial version

eg
=FACT(100)/(2*FACT(100-2))

or
=COMBIN(100,2)

Now i think about it, it all makes sense. THe number of connections is the number of ways you can choose two nodes from the whole set.

Thanks all for your input and helping me get to an answer quickly.

Are you asking a question or testing us? If you already did your homework.
Why you need to send close request? right away accept Shaun's solution and it will close automatically.

In a network with N nodes, what's the maximum number of connections you can make?

 eg N=2, gives 1 connection, N=3 gives 3, N=4 gives 6, N=5 gives 10, N=6 gives lots etc

 I think it's sum of N-i from i=1 to N-1

Consider this:
Starting with "n" nodes, each node added gets N nodes.
Each node added, adds "n" connections.
So the formula is a sum of added connections.
The number of connections is a sequence: 1,2,3,6,12,
1 to 2 adds 1 for 1 total
2 to 3 adds 2 for 3 total
3 to 4 adds 3 for 6 total
4 to 5 adds 4 for 10 total
99 to 100 adds 99 >> so we will call M=N-1 as the sequence ends at N-1
SUM over k=1:M (1:k-1) which is done with a well-known formula M*(M+1)/2
Check:
M=1>>1*(1+1)/2=1
M=2>>2*(2+1)/2=3
M=3>>3*(3+1)/2=6
M=4>>4*(4+1)/2=10
.
M=99>>99*(99+1)/2=4,950

In this case, you have to be careful because there are N nodes and N-1 added connections.  Thus M=N-1

How many when N=100?

4,950

Can I simplify this in terms of N only

Yes. As above.

what's the function called in maths

Partial sum of integers
and/or in Excel or google sheets? SUM

Incomplete answer

Thanks Fred, that's a good point. I will request to defer closing so I can re-assign points.

Nice formula btw, i see now how it's equivalent to the general case attached.

I rather hate to do this but it seems technically proper if not socially proper....