Problem
ALG-B1-M05-P006 Shifting a Recurrence
#6
★★☆☆☆ Level 2 of 5
Let \(a_1=2\), \(a_{n+1}=3a_n+4\). Find \(a_n\).
Choose \(c\) so that \(a_{n+1}+c=3(a_n+c)\).
We need \(3c=4+c\), so \(c=2\). Then \(a_{n+1}+2=3(a_n+2)\).
\(a_1+2=4\), hence \(a_n+2=4\cdot3^{n-1}\), and \(a_n=4\cdot3^{n-1}-2\).
The shift removes the constant term.