Question

Which explicit formula generates the infinite sequence 2, 9, 28, 65, 126, ...?

Answer 1

1. The terms of a sequence are denoted by $u_{1} , u_{2}, u_{3}, u_{4}, u_{5},...$

2.
 $u_{1} = 2 = 2*1 u_{2} = 9 = 3*3 u_{3} = 28 = 4*7 u_{4} = 65 = 5*13 u_{5} = 126 = 6*21$

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 $t_{1} , t_{2}, t_{3}, t_{4}, t_{5},...$:

$t_{1}=1 t_{2}=1+2 t_{3}=1+2+4=1+2+2*2 t_{4}=1+2+4+6=1+2+2*2+2*3=1+2(1+2+3)$

$t_{5}=1+2+2*2+2*3+2*4=1+2(1+2+3+4)$

5.
So

$u_{n}=(n+1)(1+2{1+2+3+....(n-1)}) =(n+1)(1+2 [(n-1)n/2]) = (n+1)(1+(n-1)n) =(n+1)( n^{2}-n+1 )$

6. We can check: $u_{3} = (3+1)( 3^{2}-3+1 )=4*(9-3+1)=4*7=28$

7. Remark: Gauss addition formula: 1+2+3+....+n=n(n+1)/2

Answer 2

Answer:

n^3+1

Step-by-step explanation:

1^3=1+1=2

2^3=8+1=9