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Algebra Level 3

( 1 − x ) ( 1 + x + x 2 + x 3 + x 4 ) = 31 32 (1-x)(1+x+x^2+x^3+x^4)=\dfrac{31}{32}

Given that x x is a rational number satisfying the equation above, find 1 + x + x 2 + x 3 + x 4 + x 5 1+x+x^2+x^3+x^4+x^5 .

63 32 \frac{63}{32} 31 64 \frac{31}{64} 31 32 \frac{31}{32} 63 64 \frac{63}{64}

Ayush G Rai by Ayush G Rai , 20, India

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1 solution

Hung Woei Neoh
Jun 24, 2016

( 1 − x ) ( 1 + x + x 2 + x 3 + x 4 ) = 31 32 1 − x + x − x 2 + x 2 − x 3 + x 3 − x 4 + x 4 − x 5 = 31 32 1 − x 5 = 31 32 x 5 = 32 32 − 31 32 = 1 32 x = 1 32 5 = 1 2 (1-x)(1+x+x^2+x^3+x^4) = \dfrac{31}{32}\\ 1\color{#3D99F6}{-x+x}\color{#D61F06}{-x^2+x^2}\color{#20A900}{-x^3+x^3}\color{#EC7300}{-x^4+x^4}-x^5 = \dfrac{31}{32}\\ 1-x^5=\dfrac{31}{32}\\ x^5 = \dfrac{32}{32} - \dfrac{31}{32} = \dfrac{1}{32}\\ x = \sqrt[5]{\dfrac{1}{32}} = \dfrac{1}{2}

Substitute the value of x x into the equation:

( 1 − 1 2 ) ( 1 + x + x 2 + x 3 + x 4 ) = 31 32 1 2 ( 1 + x + x 2 + x 3 + x 4 ) = 31 32 1 + x + x 2 + x 3 + x 4 = 62 32 (1-\dfrac{1}{2})(1+x+x^2+x^3+x^4) = \dfrac{31}{32}\\ \dfrac{1}{2}(1+x+x^2+x^3+x^4) = \dfrac{31}{32}\\ 1+x+x^2+x^3+x^4 = \dfrac{62}{32}

Add x 5 x^5 into the equation:

1 + x + x 2 + x 3 + x 4 + x 5 = 62 32 + 1 32 = 63 32 1+x+x^2+x^3+x^4+x^5 = \dfrac{62}{32}+\dfrac{1}{32} = \boxed{\dfrac{63}{32}}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution. +1

A Former Brilliant Member - 4 years, 11 months ago

Nice solution.. but instead of multiplying the brackets, you can use 1 + x + x 2 + x 3 + x 4 = 1 − x 5 1 − x \text{Nice solution.. but instead of multiplying the brackets, you can use } \\ 1+x+x^2+x^3+x^4 = \dfrac{1-x^5}{1-x}

Sabhrant Sachan - 4 years, 11 months ago

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