Fractional differences

Algebra Level 3

( 1 2 + 1 3 + . . . + 1 2005 ) × ( 1 + 1 2 + . . . + 1 2004 ) − ( 1 + 1 2 + . . . + 1 2005 ) × ( 1 2 + 1 3 + . . . + 1 2004 ) = ? \small \color{#D61F06}{ (\frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{2005}) \times ( 1 + \frac{1}{2} + . . . + \frac{1}{2004}) - (1 + \frac{1}{2} + . . . + \frac{1}{2005}) \times (\frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{2004}) = \ ?}

State your answer as numerator denominator, for example for 2 5 \frac{2}{5} , you would write 25


The answer is 12005.

Yellow Tomato by Yellow Tomato , 17, USA

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3 solutions

Kay Xspre
Oct 25, 2015

Define the following α = 1 2 + 1 3 + ⋯ + 1 2005 ; β = 1 2 + 1 3 + ⋯ + 1 2004 \alpha = \frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2005};\: \beta= \frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2004} Now just substitute the variable in the question, which gives α ( 1 + β ) − ( 1 + α ) β = α − β = 1 2005 \alpha(1+\beta)-(1+\alpha)\beta = \alpha-\beta = \frac{1}{2005}

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simple way to do this, much easier than my way XD

Yellow Tomato - 5 years, 7 months ago
Tan Chee Wen
Oct 27, 2015

Let x = 1 2 + 1 3 + . . . + 1 2004 x=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004} . Then we have ( x + 1 2005 ) ( 1 + x ) − ( 1 + x + 1 2005 ) ( x ) (x+\frac{1}{2005})(1+x)-(1+x+\frac{1}{2005})(x) = x + x 2 + 1 2005 + x 2005 − x − x 2 − x 2005 =x+x^2+\frac{1}{2005}+\frac{x}{2005}-x-x^2-\frac{x}{2005} = 1 2005 =\boxed{\frac{1}{2005}}

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Yellow Tomato
Oct 25, 2015

In order to find ( 1 2 + 1 3 + . . . + 1 2005 ) × ( 1 + 1 2 + . . . + 1 2004 ) − ( 1 + 1 2 + . . . + 1 2005 × ( 1 2 + 1 3 + . . . + 1 2004 ) = ? \small \color{teal}{ (\frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{2005}) \times ( 1 + \frac{1}{2} + . . . + \frac{1}{2004}) - (1 + \frac{1}{2} + . . . + \frac{1}{2005} \times (\frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{2004}) = \ ?} ,

all we need to do is simplify

Simplifying ( 1 2 + 1 3 + . . . + 1 2005 ) × ( 1 + 1 2 + . . . + 1 2004 ) − ( 1 + 1 2 + . . . + 1 2005 ) × ( 1 2 + 1 3 + . . . + 1 2004 ) \small \color{teal}{(\frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{2005}) \times ( 1 + \frac{1}{2} + . . . + \frac{1}{2004}) - (1 + \frac{1}{2} + . . . + \frac{1}{2005}) \times (\frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{2004}}) ,

⇒ ( 1 + 1 2 + 1 3 + . . . + 1 2005 − 1 ) × ( 1 + 1 2 + . . . + 1 2004 ) − ( 1 + 1 2 + . . . + 1 2005 ) × ( 1 + 1 2 + 1 3 + . . . + 1 2004 − 1 ) \Rightarrow \small \color{teal}{ (1 + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{2005} - 1) \times ( 1 + \frac{1}{2} + . . . + \frac{1}{2004}) - (1 + \frac{1}{2} + . . . + \frac{1}{2005}) \times ( 1 + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{2004} - 1)}

⇒ ( 1 + 1 2 + 1 3 + . . . + 1 2005 ) × ( 1 + 1 2 + . . . + 1 2004 ) − ( 1 + 1 2 + . . . + 1 2005 ) × ( 1 + 1 2 + 1 3 + . . . + 1 2004 ) − ( 1 + 1 2 + . . . + 1 2004 ) − [ − ( 1 + 1 2 + . . . + 1 2005 ) ] \Rightarrow \small \color{teal}{ (1 + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{2005} ) \times ( 1 + \frac{1}{2} + . . . + \frac{1}{2004}) - (1 + \frac{1}{2} + . . . + \frac{1}{2005}) \times ( 1 + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{2004}) - ( 1 + \frac{1}{2} + . . . + \frac{1}{2004}) - [-( 1 + \frac{1}{2} + . . . + \frac{1}{2005})]}

⇒ − ( 1 + 1 2 + . . . + 1 2004 ) + ( 1 + 1 2 + . . . + 1 2005 ) \Rightarrow \small \color{teal}{- ( 1 + \frac{1}{2} + . . . + \frac{1}{2004}) + ( 1 + \frac{1}{2} + . . . + \frac{1}{2005}})

⇒ 1 2005 \Rightarrow \huge \color{teal}{\frac{1}{2005}}

Answer is numerator (1) denominator (2005)

Which is 12005 \color{teal}{\boxed{\boxed{\boxed{\boxed{\boxed{12005}}}}}}

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