Mathematical induction?

OK, found the error:

The sum you have defined is sum(i=1 to n)i(n-i+1) <== Note "+1"

1. Let n = 1.
Then 1(1) = 1
and (1/6)(1)(1+1)(1+2) = 1.

2. Let this be true for n.
Then sum(i=1 to n)i(n-i) = (1/6)(n)(n+1)(n+2)

So for sum(i=1 to n+1)(i(n+1-i+1)) <== Note "+1"
= sum(i=1 to n+1)(i(n-i+1)) + sum(i=1 to n+1)(i)
= sum(i=1 to n)(i(n-i+1) + 0 + sum(i=1 to n+1)(i)

2a) = n(n+1)(n+2)/6 + (n+1)(n+2)/2
= (n+1)(n+2)/6 [ n + 3 ]
= (n+1)(n+2)(n+3) / 6 ==> proof since this is what you expect from the theorem.

Edit: Missing the +1 in the original summation is the issue.

The sum you have defined is sum(i=1 to n)i(n-i)

1. Let n = 1.
Then 1(1) = 1
and (1/6)(1)(1+1)(1+2) = 1.

2. Let this be true for n.
Then sum(i=1 to n)i(n-i) = (1/6)(n)(n+1)(n+2)

So for sum(i=1 to n+1)(i(n+1-i))
= sum(i=1 to n+1)(i(n-i)) + sum(i=1 to n+1)(i)
= sum(i=1 to n)(i(n-i)) + (n+1)(n-(n+1)) + sum(i=1 to n+1)(i)
= sum(i=1 to n)(i(n-i)) - (n+1) + (n+1)(n+1+1)/2
(This uses sum(i=1 to n)i = n(n+1)/2.)
= (1/6)(n)(n+1)(n+2) - (n+1) + (n+1)(n+2)/2 (induction step)
= (n+1)((1/6)(n)(n+2) - 1 + (1/2)(n+2))
= (n+1)(n^2/6 + n/3 - 1 + n/2 + 1)
= (n+1)(n^2/6 + 5n/6)
= (1/6)(n+1)(n)(n+5)

I must be making a mistake somewhere. I'll keep looking.

edit: namenot, I still don't see where the n+1 term in the first sum went. I think it should still be there.

edit (again): Thanks for the comment, namenot. I am not seeing the problem either.

edit (again again): Ah, good call. My bad!!

s6=21n-70
s7=28n-112
sn=n(n+1)/2-2n(n+1)(n+2)