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Algebra Level 5

Let f ( x ) = x 3 1 − 3 x + 3 x 2 f(x) = \frac {x^3}{1-3x+3x^2}

If the value of the expression below can be written in the form of a b \frac {a}{b} for coprime positive integers a , b a,b , what is the value of a + b a+b ?

f ( 1 2014 ) + f ( 2 2014 ) + … + f ( 2013 2014 ) f \left ( \frac {1}{2014} \right ) + f \left ( \frac {2}{2014} \right ) + \ldots + f \left ( \frac {2013}{2014} \right )


The answer is 2015.

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Thanh Viet by Thanh Viet , 21, Vietnam

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2 solutions

Aareyan Manzoor
Dec 24, 2014

f ( x ) = x 3 1 − 3 x + 3 x 2 = x 3 ( 1 − x ) 3 + x 3 f(x)=\dfrac{x^3}{1-3x+3x^2} =\dfrac{x^3}{(1-x)^3+x^3} f ( 1 − x ) = ( 1 − x ) 3 ( 1 − ( 1 − x ) ) 3 + ( 1 − x ) 3 = ( 1 − x ) 3 x 3 + ( 1 − x ) 3 f(1-x)=\dfrac{(1-x)^3}{(1-(1-x))^3+(1-x)^3}=\dfrac{(1-x)^3}{x^3 +(1-x)^3} f ( x ) + f ( 1 − x ) = x 3 ( 1 − x ) 3 + x 3 + ( 1 − x ) 3 x 3 + ( 1 − x ) 3 = x 3 + ( 1 − x ) 3 x 3 + ( 1 − x ) 3 = 1 f(x)+f(1-x)=\dfrac{x^3}{(1-x)^3+x^3}+\dfrac{(1-x)^3}{x^3 +(1-x)^3}=\dfrac{x^3 +(1-x)^3}{x^3 +(1-x)^3}=1 there are 1006 such pairs, but 1007 2014 \dfrac{1007}{2014} or 1 2 \dfrac{1}{2} is left, compute f ( 1 2 ) = ( 1 2 ) 3 2 ( 1 2 ) 3 = 1 2 f(\dfrac{1}{2})=\dfrac{(\dfrac{1}{2})^3}{2(\dfrac{1}{2})^3}=\dfrac{1}{2} now add all ( 1006 × 1 ) + 1 2 = 1006 + 1 2 = 2013 2 (1006 \times 1 )+\dfrac{1}{2}=1006+\dfrac{1}{2}=\boxed{\dfrac{2013}{2}} and 2013 + 2 = 2015 = H a p p y N e w Y e a r ! 2013+2=\boxed{2015}=\boxed{\LARGE{\color{#20A900}{H}\color{#D61F06}{a}\color{#3D99F6}{pp}\color{#69047E}{y}\quad \color{#20A900}{N}\color{#D61F06}{e}\color{#3D99F6}{w}\quad\color{#20A900}{Y}\color{#D61F06}{e}\color{#3D99F6}{a}\color{#69047E}{r}\color{#624F41}{!}}}

15 Helpful 0 Interesting 0 Brilliant 0 Confused

I did the same way :)

Ahmed Arup Shihab - 6 years, 4 months ago

Fantastic solution!

Sriram Vudayagiri - 6 years, 4 months ago
Sujoy Roy
Dec 22, 2014

Given f ( x ) = x 3 1 − 3 x + 3 x 2 = x 3 1 − 3 x ( 1 − x ) f(x)=\frac{x^3}{1-3x+3x^2}=\frac{x^3}{1-3x(1-x)} .

Therefore f ( x ) + f ( 1 − x ) = x 3 1 − 3 x ( 1 − x ) + ( 1 − x ) 3 1 − 3 x ( 1 − x ) = f(x)+f(1-x) = \frac{x^3}{1-3x(1-x)}+\frac{(1-x)^3}{1-3x(1-x)}= = x 3 + ( 1 − x ) 3 1 − 3 x ( 1 − x ) = 1 =\frac{x^3+(1-x)^3}{1-3x(1-x)}=1 .

Now, A = ∑ i = 1 2013 f ( i 2014 ) A=\sum^{2013}_{i=1}f(\frac{i}{2014})

= ∑ i = 1 1006 [ f ( i 2014 ) + f ( 2014 − i 2014 ) ] + f ( 1007 2014 ) =\sum^{1006}_{i=1}[f(\frac{i}{2014})+f(\frac{2014-i}{2014})]+f(\frac{1007}{2014})

= ∑ i = 1 1006 [ f ( i 2014 ) + f ( 1 − i 2014 ) ] + f ( 1 2 ) =\sum^{1006}_{i=1}[f(\frac{i}{2014})+f(1-\frac{i}{2014})]+f(\frac{1}{2})

= 1006 + 1 2 = 2013 2 =1006+\frac{1}{2}=\frac{2013}{2}

So, a + b = 2013 + 2 = 2015 a+b=2013+2=\boxed{2015}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

The same way as mine :))

Thanh Viet - 6 years, 5 months ago

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Please change the second term in the question to f ( 2 2014 ) \Large{ f( \frac{2}{2014}) }

Department 8 - 5 years, 6 months ago

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