Nested Cuberoots

Algebra Level 5

Compute:

54 + 972 + 21870 + 551124 + 14526054 + 3 3 3 3 3 \sqrt[3]{54+\sqrt[3]{972+\sqrt[3]{21870+\sqrt[3]{551124+\sqrt[3]{14526054+\cdots}}}}}

Details and assumptions: n th n^\text{th} term is given by 2 7 n + 3 2 n + 1 27^n + 3^{2n+1} .


The answer is 4.

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1 solution

Rishabh Jain
Jan 30, 2016

Just loved this question.. :D x + 1 = ( x + 1 ) 3 3 x+1=\sqrt[3]{(x+1)^3} = x 3 + 3 x 2 + 3 x + 1 3 =\sqrt[3]{x^3+3x^2+\color{#D61F06}{3x+1}} = x 3 + 3 x 2 + ( 3 x + 1 ) 3 3 3 =\sqrt[3]{x^3+3x^2+\sqrt[3]{\color{#D61F06}{(3x+1)^3}}} = x 3 + 3 x 2 + 27 x 3 + 27 x 2 + 9 x + 1 3 3 =\sqrt[3]{x^3+3x^2+\sqrt[3]{27x^3+27x^2+\color{#20A900}{9x+1}}} = x 3 + 3 x 2 + 27 x 3 + 27 x 2 + ( 9 x + 1 ) 3 3 3 =\sqrt[3]{x^3+3x^2+\sqrt[3]{27x^3+27x^2+\color{#20A900}{(9x+1)^3}}} = x 3 + 3 x 2 + 27 x 3 + 27 x 2 + 729 x 3 + 243 x 2 + 27 x + 1 3 3 3 =\sqrt[3]{x^3+3x^2+\sqrt[3]{27x^3+27x^2+\sqrt[3]{729x^3+243x^2+\color{#3D99F6}{27x+1}}}} = =\cdots Put x=3 to get the desired result i.e 4 = 54 + 972 + 21870 + 551124 + 14526054 + 3 3 3 3 3 \Large\color{#624F41}{\boxed{\color{#69047E}{4}}}=\sqrt[3]{54+\sqrt[3]{972+\sqrt[3]{21870+\sqrt[3]{551124+\sqrt[3]{14526054+\cdots}}}}}

Tremendously incredible approach

tanay gaurav - 5 years, 4 months ago

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Thanks! \large \color{#D61F06}{\text{Thanks!}}

Rishabh Jain - 5 years, 4 months ago

Always awesome solutions!!!

Racchit Jain - 5 years, 4 months ago

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Yup.... I loved writing a solution for this problem..... BTW Thanks!! \large\color{darkviolet}{\text{Thanks!!}}

Rishabh Jain - 5 years, 4 months ago

Bro are you a wizard???? How the hell you thought of such a wonderful approach??

Surya Sharma - 5 years, 4 months ago

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Nope... Just observation and a little mathematical intuition... :-}

Rishabh Jain - 5 years, 4 months ago

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