Question

Please show all work:

Let P₁ = 1

If x is odd then P_x+1 = 2P_x

If x is even then P_x+1 = 2P_x +1   

Prove that 2P_x+1 + 2P_x+1 +1 = P_x+2 is true and then solve it.

Answer 1

First note that

  $P_1=1\\so\ P_2=P_{1+1}=2P_1=2\ \&\\ P_3=P_{2+1}=2P_2+1=2\times2+1=5$

But if the given relation were true then for x=1, we should have$2P_2+2P_2+1=P_3\\ i.e.,\ 2\times2+2\times2+1=5,\ \text{which is impossible}$.

Rather by definition of P, it will satisfy   $P_{x+2}=P_{x+1}+2P_x+1,\ \forall x\geq 1$. Let us prove this.

If x=2k, even, then

$\begin{align*} P_{x+1}+2P_x+1=&P_{2k+1}+2P_{2k}+1\\ =&P_{2k+1}+P_{2k+1}\\ =&2P_{2k+1}\\ =&P_{2k+2}=P_{x+2} \end{align*}$.

If x=2k-1, odd, then

  $\begin{align*} P_{x+1}+2P_x+1=&P_{2k}+2P_{2k-1}+1\\ =&P_{2k}+P_{2k}+1\\ =&2P_{2k}+1\\ =&P_{2k+1}=P_{x+2} \end{align*}$.

Now to solve the relation.

$\begin{align*} For\ x=2k-1,odd\ P_{2k}=P_{x+1}=&2P_x\\ =&2P_{2k-1}\\ =&2(2P_{2k-2}+1)\\ =&2^2P_{2k-2}+2\\ =&2^3P_{2k-3}+2\\ =&...\\ =&2^{2k-1}P_1+2^{2k-3}+...+2^1\\ =&2+2^3+...+2^{2k-1}\\ =&2\frac{(2^2)^k-1}{2^2-1}\\ =&\frac{2}{3}(4^k-1)\\ =&\frac{2}{3}(2^{x+1}-1) \end{align*}$

$\begin{align*} For\ x=2k,even\ P_{2k+1}=P_{x+1}=&2P_x+1\\ =&2P_{2k}+1\\ =&2^2P_{2k-1}+1\\ =&2^2(2P_{2k-2}+1)+1\\ =&2^3P_{2k-2}+2^2+1\\ =&2^4P_{2k-3}+2^2+1\\ =&...\\ =&2^{2k}P_1+2^{2k-2}+...+2^0\\ =&1+2^2+...+2^{2k}\\ =&1\frac{(2^2)^{k+1}-1}{2^2-1}\\ =&\frac{1}{3}(2^{2k+2}-1)\\ =&\frac{1}{3}(2^{x+2}-1) \end{align*}$

Hence

$P_x=\frac{2}{3}(2^x-1)\text{ if }x\text{ is even and}\\ P_x=\frac{1}{3}(2^{x+1}-1)\text{ if }x\text{ is odd}$