Question

\(\dfrac{x^9+x^8+x^7+...+1}{x^2-1}\)=?

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Answer 1

\(\dfrac{x^9+x^8+x^7+...+1}{x^2-1}=\dfrac{\left(x^9+x^8\right)+\left(x^7+x^6\right)+...+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^8\left(x+1\right)+x^6\left(x+1\right)+...+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x^8+x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^8+x^6+x^4+x^2+1}{x-1}\)