Inductive proof that the total sum of all of the possible two element subset sums of the first n positive integers is given by .... (n + 1)C(n, 2)
Show that it is true for the first two positive integers
(2 + 1)C(2,2) = 3(1) = 3
Asuume it's true for n = k....that is......
(k + 1)C(k, 2) = (k+1) [k!]/[ (k - 2)! 2!] = (k-1)(k)(k+1) / 2!
Prove it's true for n = k + 1 ....that is...(k+1)C(k+1, 2) = (k)(k+1)(k+2) /2!
So we have
[(k + 1) + 1)] C(k+1, 2) =
(k + 1)C(k+1, 2) + C(k+1, 2) =
(k + 1)[(k + 1)!] / [(k-1)!2!] + [(k + 1)!] / [(k-1)!2!] =
(k + 1)[ (k+1)(k)] /2! + [(k+1)(k) /2!] =
[(k + 1)(k) /2!] [ (k+1) + 1) ] =
[(k + 1)(k) /2!] [ (k+2) ] =
(k)(k+1)(k+2) / 2! ......which is what we wished to show....
