Bit confused on how this:
T6+2T6+3T6+...+(n-1)T6+nT6
factored gives you:
T6[1/2(n^2+n)]
Thanks
I'm not sure what the notation T6 means, but it looks like you can divide it out of the equation, giving:
1+2+3+...+(n-1)+n = 1/2(n^2+n) (which is equal to n(n+1)/2)
If n is even, you can group the left hand side like this:
(1+n) + (2+(n-1)) + (3+(n-2)) + ... + ((n/2)+((n/2)+1)
and there are n/2 terms which are all equal to n+1, so the sum is n(n+1)/2.
If n is odd, you can do the same thing without the last term:
1+2+3+...+(n-1)+n = (1+2+3+...+(n-1)) + n = (n-1)(n)/2 + n
which simplifies to (n^2-n)/2 + 2n/2 = (n^2-n+2n)/2 = (n^2+n)/2 = n(n+1)/2