Question

(1)+(1+1)+(1+1+1)+.........+(1+1+1+......n-1 times)= ?

Answer 1

`S=n*(n-1)/2`

Explanation:

So, this sequence boils down to:

1+2+3+4+5+6+7+8+ ... + (n-1), for any given n, so if n were 1000 we would go up to 999.

Now, if we break that sequence up to half of our final member and flip each half onto the other we get this (as an example if our n was 11):

1 + 2 + 3 + 4 + 5 +
10+9 + 8 + 7 +6

Notice that 10 + 1 = 11, and 2 + 9 = 11, and 3 +8 = 11... so on and so forth. So the sum of the first member and the last, and each other pair is equal to our initial n. Now count how many times we did this pairing. In this case it was 5 for our n of 11 or (11-1)/2. Let's try and n of 14, to see what happens:

1 + 2 + 3 + 4 + 5 + 6 + 7 +
13+12+11+10+9 + 8

We now have totals of 14 (our n again), and that 7 hanging there without a pair. Notice that 7 is exactly 14/2, so it is as if we did half an addition, or 6.5 additions of 14. Notice that 6.5=(14-1)/2, or (n-1)/2

Now we have enough information and generalizations to make our equation. For whatever n we get, we sum n, (n-1)/2 times, giving us:

`S=n*(n-1)/2`