Stuck on solving recurrence relation

Question

I'm trying to find formula for the following sequence.

1, 3, 6, 10, 15...

Recursive formula is pretty straightforward

My attempt to solve it:

1. Homogeneous solution

   

   

   

   

2. Particular solution

   

   

   

I stuck here. A bit of help will be appreciated.

Thanks in advance.

Answer 1

As $a_n-a_{n-1}=n$

$a_n$ can be at most quadratic $a_n=An^2+Bn+C$ where $A,B,C$ are arbitrary constants

Now $a_{n-1}=A(n-1)^2+B(n-1)+C$

So, $An^2+Bn+C-\{A(n-1)^2+B(n-1)+C\}=n$

$\implies 2An-(A-B)=n$

Equating the different powers of $n$ we get $A-B=0$ and $2A=1$

So, $a_n=\frac{n^2}2+\frac n2+C$

For $n=1=a_1=\frac12+\frac12+C\implies C=0$

Answer 2

Try $a_n=An^2+Bn$ for a particular solution.