Hello, CPR!
Recurrence relation
. . $\displaystyle d_1 \:=\:1,\quad d_2\:=\:4,\quad d_n\:=\:2d_{n-1} - d_{n-2}$
I would immediately crank out a few terms . . .
. . $\displaystyle \begin{array}{ccccc} d_1 &=& 1 \\ d_2 &=& 4 \\ d_3 &=& 2(4)-1 &=& 7 \\ d_4 &=& 2(7)-4 &=& 10 \\ d_5 &=& 2(10)-7 &=& 13 \\ d_6 &=& 2(13)-10 &=& 16 \\ \vdots & & \vdots & & \vdots
\end{array}$
We have an arithmetic sequence: first term 1, common difference 3.
The general term is: . $\displaystyle d_n \;=\;3n-2$
Thanks Soroban! However, I was wondering if
dn = d1 (subscript didnt work)
Since 1 is the first term and the common difference is 3, do we simply say 4-1
then n=1
How do we get dn = 3n-2, is really what Im asking?
How do I do subscripts so they will show up here?