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Omni-MATH / Let f(x)=x²+2x+1. Let g(x)=f(f(⋯f(x))), where there are 2009 f mathrm{~s} in the expression for g(x). Then g(x) can be written as g(x)=x^(2²⁰⁰⁹)+a_(2²⁰⁰⁹−1)x^(2²⁰⁰⁹−1)+⋯+a₁x+a₀ whe…

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problem

Let f(x)=x2+2x+1f(x)=x^{2}+2 x+1. Let g(x)=f(f(⋯f(x)))g(x)=f(f(\cdots f(x))), where there are $2009 f \mathrm{~s}$ in the expression for g(x)g(x). Then g(x)g(x) can be written as g(x)=x22009+a22009−1x22009−1+⋯+a1x+a0g(x)=x^{2^{2009}}+a_{2^{2009}-1} x^{2^{2009}-1}+\cdots+a_{1} x+a_{0} where the aia_{i} are constants. Compute a22009−1a_{2^{2009}-1}.
Plain-text mathematical notation (without MathML)
Let f(x)=x²+2x+1. Let g(x)=f(f(⋯f(x))), where there are $2009 f \mathrm{~s}$ in the expression for g(x). Then g(x) can be written as g(x)=x^(2²⁰⁰⁹)+a_(2²⁰⁰⁹−1)x^(2²⁰⁰⁹−1)+⋯+a₁x+a₀ where the a_(i) are constants. Compute a_(2²⁰⁰⁹−1).
Original LaTeX notation
Let $f(x)=x^{2}+2 x+1$. Let $g(x)=f(f(\cdots f(x)))$, where there are $2009 f \mathrm{~s}$ in the expression for $g(x)$. Then $g(x)$ can be written as $g(x)=x^{2^{2009}}+a_{2^{2009}-1} x^{2^{2009}-1}+\cdots+a_{1} x+a_{0}$ where the $a_{i}$ are constants. Compute $a_{2^{2009}-1}$.

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task_id
aafec071-b493-5dc2-9d20-2671516c67cb
task_key
test--aafec071-b493-5dc2-9d20-2671516c67cb
task_revision_id
1
config / split
/ test

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