How to solve the recurrence relation $a_n – 2 a_{n-1} = 3 \times 2^n, a_0 = 1$. By looking at the terms of the relation, it can be seen that it is linear in nature but it is not homogeneous. How to solve such a recurrence relation?
[Math] How to solve the recurrence relation $a_n – 2 a_{n-1} = 3 \times 2^n, a_0 = 1$
algebra-precalculusrecurrence-relations
Best Answer
Hint: let $\,a_n=2^n b_n\,$, then $\,2^n b_n - 2 \cdot 2^{n-1} b_{n-1} = 3 \cdot 2^{n} \iff b_n - b_{n-1}=3\,$, so $b_n$ is an arithmetic progression, with common difference $\,3\,$ and $\,b_0=a_0 / 2^0 = 1\,$.