Question

Prove by induction:

Summation where j=1 to n-1: j

equals (n(n-1)/2)

Answer 1

\[\sum_{j=1}^{n-1}j=\frac{ n(n-1) }{2 }\]

Answer 2

Here is what I have, but it seems way to elementary

Answer 3

Assume j=2. We gather that 2(2-1)/2=2 thus making the equation true.

Now we will assume that for all k which will be equal to (n-1) now prove that it is also true for k+1.

Answer 4

\[\frac{ (k+2)(k+1) }{ 2 }=\frac{ k^2+3k+2 }{ 2 }\]

Answer 5

If it works for n then you have for n+1:
\[\sum_{k=1}^{n+1}k=\sum_{k=1}^{n-1}k + (n+1)+n = \frac{n(n-1)}{2}+2n+1 =\frac{n^2-n+4n+2}{2}=\frac{(n+1)(n+2)}{2}\]
As you can rewrite:
\[\sum_1^{n-1}k =\frac{n(n-1)}{2} \implies \sum_{k=1}^{n}\frac{n(n+1)}{2}\]
So this transforms into:
\[\frac{n+1((n+1)+1)}{2}\]
Which is what you want.

Answer 6

so in essence I was right I just had to separate the k+2, got it! thanks