# Thread: Showing LHS = RHS

1. ## Showing LHS = RHS

Ok, i have go to a point working through a question where i am having to show that the LHS = RHS but cant seem to.... is the following possible or have i gone wrong somewhere earlier in my calculations?

If P(Ak) = a + b* $({q/p})^k$

Then use the equivalent formula for P(Ak+1) and P(Ak-1) and show that....

P(Ak+1) =
$\frac{P(Ak) - q*P(Ak-1)}{p}$

2. Originally Posted by sirellwood
Ok, i have go to a point working through a question where i am having to show that the LHS = RHS but cant seem to.... is the following possible or have i gone wrong somewhere earlier in my calculations?

If P(Ak) = a + b* $({q/p})^k$

Then use the equivalent formula for P(Ak+1) and P(Ak-1) and show that....

P(Ak+1) =
$\frac{P(Ak) - q*P(Ak-1)}{p}$