Solve the following recurrence by using telescoping

Question

a(n)=2a(n-1) + 2n-1

a(0)=1

I tried below;

a(n)-2a(n-1)=2n-1

from here I found

P(n)=1, q(n)=2 r(n)=? according to below formula

p(n)an()-q(n)a(n-1)=r(n) for n>=1

Since I can not find r(n) value, I can not solve it, any help appreciated. Thanks in advance

Answer 1

Let $a(n)=b(n)+An+B$

$a(n)-2a(n-1)=b(n)+An+B-2[b(n-1)+A(n-1)+B]$

$\iff2n-1=b(n)-2b(n-1)-An+2A-B$

Set $-A=2,2A-B=-1$ to find $b(n)=2b(n-1)=\cdots=2^r\cdot b(n-r)$ where $r$ is any integer