$a_n$ = $a_{n-1} + 3n – 5$ $,$ $a_0 = 7$
So far what I got is:
$a_0 = 7$
$a_1 = 7+3(1)-5$
$a_2 = 7+3(1)-5 + 3(2)-5$
$a_3 = 7+3(1)-5 + 3(2)-5 + 3(3)-5$
$a_4 = 7+3(1)-5 + 3(2)-5 + 3(3)-5 + 3(4)-5$
This is where I'm stuck I'm having trouble finding a pattern to the relation.
At first I tried solving this as a quadratic sequence and got $a_n = \frac{3}{2}{n^2} – \frac{7}2n + 7$ which works but on a test I solved a recurrence relation problem as a quadratic sequence got the right answer but got a 0 on the problem because the professor wanted us to use forward or backward substitution which is what I am attempting above (forward substitution).
Best Answer
By telescoping sum, $a_n-7=a_n-a_0=\sum\limits_{k=1}^n(a_k-a_{k-1})=\sum\limits_{k=1}^n(3k-5)=3\sum\limits_{k=1}^n k-\sum\limits_{k=1}^n 5=3\frac{n(n+1)}2-5n$