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Algebra Level 4

f ( x ) = ( x + 1 ) ( x 2 − x ) x x + x + x \large f(x) = \dfrac{(\sqrt{x} + 1)(x^{2} - \sqrt{x})}{x \sqrt{x} + x + \sqrt{x}}

Define f ( x ) f(x) as above and further define g ( x ) = f ( f ( x ) ) g(x) = f(f(x)) .

Find the value of [ ∑ i = 2 1000 g ( i ) ] − 1 \large\displaystyle [\sum_{i=2}^{1000} g(i)] - 1 .


The answer is 498500.

Harsh Shrivastava by Harsh Shrivastava , 20, India

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2 solutions

f ( x ) = ( x + 1 ) ( x 2 − x ) x x + x + x = x ( x + 1 ) ( x x − 1 ) x ( x + x + 1 ) = ( x − 1 ) ( x + 1 ) ( x x − 1 ) ( x − 1 ) ( x + x + 1 ) = ( x − 1 ) ( x x − 1 ) x x − 1 = x − 1 f(x)=\frac { \left( \sqrt { x } +1 \right) \left( { x }^{ 2 }-\sqrt { x } \right) }{ x\sqrt { x } +x+\sqrt { x } } =\frac { \sqrt { x } \left( \sqrt { x } +1 \right) \left( { x }\sqrt { x } -1 \right) }{ \sqrt { x } \left( x+\sqrt { x } +1 \right) } =\frac { \left( \sqrt { x } -1 \right) \left( \sqrt { x } +1 \right) \left( { x }\sqrt { x } -1 \right) }{ \left( \sqrt { x } -1 \right) \left( x+\sqrt { x } +1 \right) } =\frac { \left( x-1 \right) \left( { x }\sqrt { x } -1 \right) }{ x\sqrt { x } -1 } =x-1\\

g ( x ) = f ( f ( x ) ) = f ( x − 1 ) = x − 2 \\ g(x)=f(f\left( x \right) )=f(x-1)=x-2\\

∑ i = 1 1000 i − 2 = ∑ i = 1 1000 i − ∑ i = 1 1000 2 = 500500 − 2000 = 498500 \\ \sum _{ i=1 }^{ 1000 }{ i-2=\sum _{ i=1 }^{ 1000 }{ i } -\sum _{ i=1 }^{ 1000 }{ 2 } } =500500-2000=498500

9 Helpful 0 Interesting 0 Brilliant 0 Confused
Mj Bltz
Jun 4, 2015

We can simplify f ( x ) f(x) as follows:

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

After cancelling the x \sqrt{x} terms, and multiplying the grouped terms in the numerator, we get

f ( x ) = x 2 − 1 + x x − x x + x + 1 = ( x + 1 ) ( x − 1 ) + ( x − 1 ) x x + x + 1 f(x) = \frac{x^2 -1+x\sqrt{x}- \sqrt{x}}{x+\sqrt{x}+1} = \frac{(x+1)(x-1)+(x-1)\sqrt{x}}{x+\sqrt{x}+1} = ( x − 1 ) [ ( x + 1 ) + x ] x + x + 1 = x − 1 = \frac{(x-1)[(x+1)+\sqrt{x}]}{x+\sqrt{x}+1} = x-1

And thus, g ( x ) = f ( f ( x ) ) = f ( x − 1 ) = ( x − 1 ) − 1 = x − 2 g(x)= f(f(x))= f(x-1)= (x-1)-1 = x-2 ∑ i = 1 1000 g ( i ) = ∑ i = 1 1000 ( i − 2 ) = ( − 1 + 998 ) ( 1000 ) 2 = 498500 \sum_{i=1}^{1000}g(i)= \sum_{i=1}^{1000}(i-2) = (-1+998)\frac{(1000)}{2} = 498 500

3 Helpful 0 Interesting 0 Brilliant 0 Confused

There was no need no expand you could have straight away used the formula for a 3 − b 3 {a}^{3}-{b}^{3} .

Abhishek Sharma - 6 years ago

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@Harsh Shrivastava I think that the summation should not start with 1 cause f(f(1)=0/0

A Former Brilliant Member - 6 years ago

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