I'm trying to solve this for b:
b*2^n = 3*b*2^(n-1) + 2^n
the answer is:
b = -2^(n + 1)
but i don't see how to get there.
can someone enlighten me?
thanks
dan mcleran
an = 3an-1 + 2^nwith a1 = 5
I was able to solve the homogeneous part to get:
f(n) = a•3^n
I then substituted the particular solution into the above recurrence:
b•2^n = 3 b•2^(n-1) + 2^n
But I think I am too rusty at the algebra to solve for b. I was really looking for some algebra help I guess.
Thanks
Dan
Hello, Dan!
As we say in the Marines, "Ya done good!"$\displaystyle a_n \:=\: 3a_{n-1} + 2^n,\quad a_1 = 5$
I was able to solve the homogeneous part to get: .$\displaystyle f(n) \:=\: a\!\cdot\!3^n$
I then substituted the particular solution into the above recurrence:
. . $\displaystyle b\cdot 2^n \:=\: 3b \cdot 2^{n-1} + 2^n$
But I think I am too rusty at the algebra to solve for $\displaystyle b.$
We have: .$\displaystyle b\!\cdot\!2^n \:=\: 3b\!\cdot\!2^{n-1} + 2^n$
Divide by $\displaystyle 2^{n-1}\!:\;\;2b \:=\:3b + 2 \quad\rightarrow\quad b \:=\:-2$
Hence: .$\displaystyle f(n) \;=\;a\!\cdot\!3^n - 2\!\cdot\!2^n $
Since $\displaystyle a_1 = 5$, we have: .$\displaystyle a\!\cdot\!3 - 2^2 \:=\:5 \quad\Rightarrow\quad a \,=\,3$
. . Hence: .$\displaystyle f(n) \;=\;3\!\cdot\!3^n - 2\!\cdot\!2^n$
Therefore: .$\displaystyle f(n) \;=\;3^{n+1} - 2^{n+1}$