Find the recurrence relation solution. $a_n$ = $a_{n-1} + 3n - 5$

Question

$a_n$ = $a_{n-1} + 3n - 5$ $,$ $a_0 = 7$

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).

Answer 1

By telescoping sum, $a_n-7=a_n-a_0=\sum\limits_{k=1}^n(a_k-a_{k-1})=\sum\limits_{k=1}^n(3k-5)=3\sum\limits_{k=1}^n k-\sum\limits_{k=1}^n 5=3\frac{n(n+1)}2-5n$

Answer 2

Observe that $$a_n - a_0 = \sum\limits_{k=1}^n (a_k - a _{k-1}) = 3 \sum\limits_{k=1}^{n} k - 5 n.$$

Answer 3

Let $A(z) = \sum_{n=0}^\infty a_nz^n$. Multiply both sides of the recurrence by $z^n$ and sum over $n$ to obtain

$$ \sum_{n=1}^\infty a_nz^n = \sum_{n=1}^\infty a_{n-1}z^n + \sum_{n=1}^\infty 3nz^n - \sum_{n=1}^\infty 5z^n. $$ By some algebra we obtain $$ A(z) - a_0 = zA(z) +\sum_{n=0}^\infty 3nz^n - \sum_{n=1}^\infty 5z^n. $$ From the identity $\sum_{n=0}^\infty z^n = \frac1{1-z}$ and differentiating we obtain $$ \sum_{n=0}^\infty 3nz^n - \sum_{n=1}^\infty 5z^n= \frac{3 z}{(1-z)^2} -\frac{5 z}{1-z}, $$ so we have $$ A(z) - 7 = zA(z) +\frac{3z}{(1-z)^2} - \frac{5z}{1-z}. $$ Solving for $A(z)$ yields $$ A(z) = \frac{12 z^2-16 z+7}{(1-z)^3}. $$ After a partial fraction decomposition, we have $$ A(z) = -\frac{8}{(1-z)^2}+\frac{3}{(1-z)^3}+\frac{12}{1-z}. $$ Writing the terms on the right-hand side as series yields \begin{align} A(z) &= \sum_{n=0}^\infty -8(n+1)z^n + \sum_{n=0}^\infty \frac32(n+1)(n+2)z^n + \sum_{n=0}^\infty 12z^n\\ &= \sum_{n=0}^\infty \left(\frac{1}{2} n (3 n-7)+7\right) z^n. \end{align} It follows then that $$ a_n = \frac{1}{2} n (3 n-7)+7. $$