1. The terms of a sequence are denoted by [tex] u_{1} , u_{2}, u_{3}, u_{4}, u_{5},... [/tex]
2.
[tex] u_{1} = 2 = 2*1
u_{2} = 9 = 3*3
u_{3} = 28 = 4*7
u_{4} = 65 = 5*13
u_{5} = 126 = 6*21
[/tex]
3. so it is clear that the first columns add each time by one, and the second column add by 2, then by 4, by 6, by 8 and so on.
4. consider only the second column and how we get the terms, which we will call [tex] t_{1} , t_{2}, t_{3}, t_{4}, t_{5},... [/tex]:
[tex]t_{1}=1
t_{2}=1+2
t_{3}=1+2+4=1+2+2*2
t_{4}=1+2+4+6=1+2+2*2+2*3=1+2(1+2+3)[/tex]
[tex]t_{5}=1+2+2*2+2*3+2*4=1+2(1+2+3+4)[/tex]
5.
So
[tex]u_{n}=(n+1)(1+2{1+2+3+....(n-1)})
=(n+1)(1+2 [(n-1)n/2])
= (n+1)(1+(n-1)n)
=(n+1)( n^{2}-n+1 )
[/tex]
6. We can check: [tex] u_{3} = (3+1)( 3^{2}-3+1 )=4*(9-3+1)=4*7=28 [/tex]
7. Remark: Gauss addition formula: 1+2+3+....+n=n(n+1)/2
