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!"
I was able to solve the homogeneous part to get: .
I then substituted the particular solution into the above recurrence:
. .
But I think I am too rusty at the algebra to solve for
We have: .
Divide by
Hence: .
Since , we have: .
. . Hence: .
Therefore: .