So we need to find three coefficients, so we will need three points:
Let's choose the first three (although ANY three will work):
n = 1 --> S(1) = 1
n = 2 --> S(2) = 1 + 2 = 3
n = 3 --> S(3) = 1 + 2 + 3 = 6
So we have:
S(1) = a * 1² + b * 1 + c = 1
S(2) = a * 2² + b * 2 + c = 3
S(3) = a * 3² + b * 3 + c = 6
--> which yields this linear equation
a + b + c = 1
4a + 2b + c = 3
9a + 3b + c = 6
let's eliminate c by subtracting it from the first:
a + b + c = 1
3a + b + 0 = 2
8a + 2b + 0 = 5
Now let's solve the bottom two:
3a + b = 2
8a + 2b = 5
--> multiply first equation by 2
6a + 2b = 4
--> subtract from second
2a + 0 = 1
-->
a = 1/2
Go back and find b
6 * 1/2 + 2b = 4
-->
b = (4 - 3) / 2 = 1/2
Now find c:
1/2 + 1/2 + c = 1
-->
c = 1 - 1 = 0
So we have:
a = b = 1/2 and c = 0:
S(n) = 1/2 * n² + 1/2 * n + 0
--> or we can factor out 1/2 * n
-->
S(n) = n/2 * (n + 1) = n(n + 1) / 2...qed