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Find out how many ways to paint a $1 × n$ chessboard with an even number of black and any number of white squares using the:

  1. exponential stretching function,
  2. using a combinatorial argument.

I tried to solve the first one this way: $$B(black)=(1,0,1,0,...)$$ $$W(white)=(1,1,1,1,...)$$ $$B(x)\cdot W(x)=((e^x+e^{-x})\cdot e^x)/2=(e^{2x}+1)/2$$

Here I don't know how to continue .

hardmath
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Lisa
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  • Welcome to MSE. Please sue MathJax to format your posts. – saulspatz Jul 05 '20 at 22:12
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    What is a "stretching function"? – bof Jul 05 '20 at 22:13
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    Your Question leaves some room for Readers to guess at what to count. Will all squares of the "chessboard" be painted either black or white? Does calling it a "chessboard" imply that colors alternate? If a pair of colorings are identical upon rotation, do we count them as one coloring or two? – hardmath Jul 05 '20 at 22:19
  • https://math.stackexchange.com/questions/248245/exactly-half-of-the-elements-of-mathcalpa-are-odd-sized – user1001001 Jul 06 '20 at 15:06
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    Lisa, you've not apparently returned to the site for a couple of days after posting this. In the circumstances you have of course not edited the Question or otherwise responded to the requests for clarification. I'm voting to close the Question until you have a chance to do that. Note that your use of $x$ in the formulas is particularly unclear, appearing as an argument to both functions $W(x),B(x)$ but without properly defining them. – hardmath Jul 08 '20 at 19:02

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