OpenStudy (anonymous):

Solve the single-term recurrence relation with variable coefficient: \(a_n = 2na_{n-1}; a_0 = 1\). I found \(a_1 = 2, a_2 = 8, a_3 = 48\), but what to do with this info? Is there a "characteristic equation" for this form?

2 years ago
ganeshie8 (ganeshie8):

@dan815

2 years ago
OpenStudy (anonymous):

What, he would know?

2 years ago
ganeshie8 (ganeshie8):

try \[a_n = 2^n n!a_0\]

2 years ago
OpenStudy (anonymous):

That does seem to work, thanks... but how did you get that?

2 years ago
ganeshie8 (ganeshie8):

\[\begin{align} a_n &=2na_{n-1}\\~\\ &=2n(2(n-1)a_{n-2}) = 2^2n(n-1)a_{n-2}\\~\\ &= 2^2n(n-1)(2(n-2)a_{n-3}) = 2^3 n(n-1)(n-2)a_{n-3}\\~\\ &=\cdots\\~\\ & = 2^nn! a_0 \end{align}\]

2 years ago