We can find a formula for nth term of the given sequence as follows:
1, 5, 12, 22, 35
The 1st differences between terms:
4, 7, 10, 13
The 2nd differences :
3, 3, 3
Since it takes two rounds of differences to arrive at a constant difference between terms, the nth term will be a 2nd degree polynomial of the form: [tex]a n^2 + b n + c[/tex], where c is a constant. The coefficients a, b, and the constant c can be found.
We can form the following 3 equations with 3 unknowns a, b, c:
[tex]1 = a\cdot1^2 + b\cdot1 + c\\5 = \cdot2^2 + b\cdot2 + c\\12 = a\cdot3^2 + b\cdot3 + c[/tex]
Solving for a, b, c, we get:
a = 3/2, b = -1/2, c = 0
Therefore, the nth term of the given sequence is:
[tex]\boxed{ a_n = \dfrac{3}{2}n^2- \dfrac{1}{2} n}[/tex]
