Adding Fractions

[MathStudent1999]

73. Calculate the following sum: 1/2*4 + 1/4*6 + 1/6*8 + ... + 1/18*20


I simplified and made it 1/8 + 1/24 + 1/48 + 1/80 + 1/120 + ... + 1/360

Now as this is a math contest question, I can't just make them all have a common denominator and then just add them. I need a fast way to do it.

I tried factoring out 1/8 but that just made it just as treacherous. What can I do to make this question simpler, easier and less time consuming?

 

[pappus]

73. Calculate the following sum: 1/2*4 + 1/4*6 + 1/6*8 + ... + 1/18*20


I simplified and made it 1/8 + 1/24 + 1/48 + 1/80 + 1/120 + ... + 1/360

Now as this is a math contest question, I can't just make them all have a common denominator and then just add them. I need a fast way to do it.

I tried factoring out 1/8 but that just made it just as treacherous. What can I do to make this question simpler, easier and less time consuming?

1. Factor out \(\displaystyle \frac18\):

\(\displaystyle \frac18 \cdot \left(\frac11 + \frac13 + \frac16 + \frac1{10} + ... + \frac1{45} \right)\)

2. You see that the denominators are elements of an arithmetic sequence of 2nd grade:

\(\displaystyle 1 + 3 + 6 + 10 + ... = \frac12 n^2 + \frac32 n + 1 = \frac12 (n+1)(n+2)\) . Start counting at zero.

3. Thus you want to evaluate:

\(\displaystyle \sum_{n=0}^8\left(\frac1{4(n+1)(n+2)} \right)\)

4. I've got the result \(\displaystyle \frac9{40}\)

 

[lookagain]

73. Calculate the following sum: 1/2*4 + 1/4*6 + 1/6*8 + ... + 1/18*20.

------> What you have is the same as (1/2)4 + (1/4)6 + ... + (1/18)20.



I tried factoring out 1/8 but that just made it just as treacherous. What can I do to make this question simpler,
easier and less time consuming?

MathStudent, you typed it wrong. You must have grouping symbols for the Order of Operations, such as


1/(2*4) + 1/(4*6) + 1/(6*8) + ... + 1/(18*20)

This doesn't fall under (just) arithmetic, because it makes use of more advanced math,
such as college algebra and series.


So, none of that algebra by pappus would you be expected to know or use here.


Factor out 1/4, as in factoring out a 2 from each factor in each denominator:


(1/4)[1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(9*10)] =


(1/4)[1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10)] ---------------> This step could be considered a "trick."



See what cancels and continue...

 

[MathStudent1999]

MathStudent, you typed it wrong. You must have grouping symbols for the Order of Operations, such as


1/(2*4) + 1/(4*6) + 1/(6*8) + ... + 1/(18*20)

This doesn't fall under (just) arithmetic, because it makes use of more advanced math,
such as college algebra and series.


So, none of that algebra by pappus would you be expected to know or use here.


Factor out 1/4, as in factoring out a 2 from each factor in each denominator:


(1/4)[1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(9*10)] =


(1/4)[1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10)] ---------------> This step could be considered a "trick."



See what cancels and continue...

All the numbers in the middle cancel out so I would be left with (1/4)[1-(1/10)] Which would equal (1/4)(9/10) which equals 9/40, which I believe is the correct answer based on the answer pappus got. Thanks all!