Let $$x=1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{...}}}};$$ then the value of $(2x-1)^2$ equals...
I don't how to start this question. Please help.
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$\begingroup$ Hint: Is there a pattern to the fraction? $\endgroup$– Dude156Feb 14, 2019 at 2:08
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2 Answers
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If you only want to know the value that the continued fraction converges to, you use a simple technique:
$$ x=1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{...}}}}=1+\frac{1}{2+\frac{1}{x}} $$
With some manipulation you could come up with the value of $x$, but you want $(2x-1)^2:$
$$ x\left(2+\frac{1}{x} \right)=\left(2+\frac{1}{x} \right)+1 $$
$$ 2x+1=3+\frac{1}{x} $$
Multiply everything by $x$, since we know it's not zero:
$$ 2x^2-2x-1=0 $$
Complete the square that we want by multiplying by $2$: $$ 4x^2-4x-2=4x^2-4x+1-3=(2x-1)^2-3=0 $$
Hence the answer is 3.
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HINT:
Write the expression as $$x=1+\frac{1}{2+\frac{1}{x}}.$$
