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Please show all work:

Let P_{1 }= 1

If x is odd then P_{x+1}= 2P_{x}

If x is even then P_{x+1}= 2P_{x} +1

Prove that P_{x+2}= 2P_{x} +1 + P_{n+1} is true and then solve it.

math   Advanced-Math   
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Answer #1

\\ Suppose \ x>2\ is\ even\\ \\ P_{x+2}=2P_{x+1}=P_{x+1}+P_{x+1}\\ \\ =2P_x +1 + P_{x+1}\\ \\ Suppose\ x>2\ is \ odd \\ \\ P_{x+2}=2P_{x+1}+1=P_{x+1}+1+P_{x+1}\\ \\ =2P_x +1 + P_{x+1}\\ \\ \therefore P_{x+2}=2P_x+1+P_{x+1}\ \ \ -(*)

\\ Take \ S(x)=P_1+xP_2+x^2P_3+...+x^{n-1}P_n+...\\ Then\ by\ (*)\ we \ get\\ S(x)(1-x-2x^2)=1+x+x^2+...+x^{n-1}+... \\ S(x)=-\frac{(1+x+...)}{(2x^2+x-1)}\\ \\=-\frac{(1+x+...)}{2(x+1)(x-\frac{1}{2})}=\frac{(1+x+x^2(1+x)+...+x^{2k}(1+x)+...)}{(x+1)(1-2x)}\\ \\=(1+x^2+x^4+...+x^{2k}+...)(1+2x+2^2x^2+2^3x^3+...)\\ \\ P_n=1+2^2+2^4+...+2^{n-1}=\frac{2^{n+1}-1}{3}\ \ \ if \ n \ is \ odd\\ \\ P_n=2+2^3+2^5+...+2^{n-1}=2\frac{2^n-1}{3}\ \ \ \ if \ n\ is\ even

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