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Find the value of \(x\) while \[\sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}+\sqrt{x^{4}+...}}}}=\sqrt{2x+1}\]

Note by Jason Snow
1 year, 9 months ago

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  Easy Math Editor

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Can you write some terms of the series in LHS ?

Aditya Sky - 1 year, 7 months ago

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Do you know the answer

Aniket Gupta - 1 year, 9 months ago

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Desmos says that \(x=4\) but I'm not sure if it's correct

Jason Snow - 1 year, 9 months ago

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\((n+1)=\sqrt { n^{ 2 }+2n+1 } =\sqrt { 1+\sqrt { n^{ 4 }+4n^{ 3 }+4n^{ 2 } } } =\sqrt { 1+\sqrt { 4{ n }^{ 3 }+\sqrt { n^{6}+8n^6+16n^4}}}\)

continuing the procedure we get

\(\sqrt { 1+\sqrt { 4n^{ 3 }+\sqrt { 16n^{ 4 }+\sqrt { 64{ n }^{ 12 }... } } } } \)

for n=1 we have

\(\sqrt { 1+\sqrt { 4+\sqrt { 16+\sqrt { 64... } } } } \)

squaring then subtracting one from each side we have above equation when \(x=4\)

we have the expression =3 when x=4

the other equation is also 3 when x=4

hence proved

Hummus A - 1 year, 9 months ago

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@Hummus A Thanks !! :D

Jason Snow - 1 year, 9 months ago

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desmos is right

i'll show you why,it will take a bit of time to type it in LaTeX

Hummus A - 1 year, 9 months ago

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We have tried some methods that didn't seem to work, so looking forward to your solution :D

Jason Snow - 1 year, 9 months ago

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