I checked the base case and it worked, but I'm not really sure what the hypothesis should be?
$$n = k, a_k < (\frac{5}{3})^k$$
If we want to prove $$k+1$$ then we have this as what we're trying to proof: $$P(k+1) = a_{k+1} < (\frac{5}{3})^{k+1}$$
I checked the base case and it worked, but I'm not really sure what the hypothesis should be?
$$n = k, a_k < (\frac{5}{3})^k$$
If we want to prove $$k+1$$ then we have this as what we're trying to proof: $$P(k+1) = a_{k+1} < (\frac{5}{3})^{k+1}$$
HINT
We have that
$$a_{k+1} = a_k + a_{k-1}\stackrel{Ind. Hyp.}<\left(\frac{5}{3}\right)^{k}+\left(\frac{5}{3}\right)^{k-1} =\left(\frac{5}{3}\right)^{k}\left(1+\left(\frac{5}{3}\right)^{-1}\right)\stackrel{?}<\left(\frac{5}{3}\right)^{k+1}$$
and the latter requires
$$\left(\frac{5}{3}\right)^{k}\left(1+\left(\frac{3}{5}\right)\right)\stackrel{?}<\left(\frac{5}{3}\right)^{k}\left(\frac{5}{3}\right)$$