To find the quadratic function t(n) which gives us the nth term tn, let t(n)=ax2+bx+c.
t(1)=a+b+c=1, t(2)=4a+2b+c=5, t(3)=9a+3b+c=13.
t(2)-t(1)=3a+b=4, t(3)-t(2)=5a+b=8,
So 2a=4, a=2, b=-2, c=1, tn=2n2-2n+1.
Sn=∑2n2-2n+1=2∑n2-2∑n+n.
∑n2=n(n+1)(2n+1)/6, ∑n=n(n+1)/2, so:
Sn=n(n+1)(2n+1)/3-n(n+1)+n=
(2n3+3n2+n-3n2-3n+3n)/3=(2n3+n)/3=⅓n(2n2+1).
Let's test it: tn=2n2-2n+1
1, 5, 13, 25, 41, ..., S5=85 (by adding the terms).
Sn=⅓n(2n2+1)=5(50+1)/3=5×17=85.