Ok.

Heres a very condensed version of my proof:

We have the question 11/15 + 11/21 + 11/28.

let a = 3, a+1 = 4, a+2 = 5, and a+4 =7.

Back in grade school we solved it like this:

11/3*5 + 11/3*7 + 11/4*7. (Now to get a common denom...)

11(4)(7)+11(4)(5)+11(3)(5) / (3)(4)(5)(7)

simplify by combining like terms

11(63) / (3)(4)(5)(7)

Now break up the 63 into its prime factorization

11(3)(3)(7) / (3)(4)(5)(7)

Now lets cross out common factors...

11(3) / (4)(5)

Now, lets substitute our original factors back into the equation:

11(a) / (a+1)(a+2)

AND THATS THE FORMULA!!!

Remember that It will only work if the numerators are all equivilant.

(Now we could have done this the long way and gotton the same answer,but I really dont feel like writing that out.)

Any dissentions?

 - Zack -

Zack, here's your counter example: a=1

1/2+1/5+1/10=4/5 when your method gives 1/6. Are you sure about that proof?



1/(a*(a+2)) + 1/(a(a+4)) + 1/((a+1)(a+4))

Time for boring algebra.

Get a common denominator.

(a+4)(a+1)+(a+2)(a+1)+(a*(a+2))
a(a+2)(a+1)(a+4)

Let's multiply polynomials! Oh golly!

(aa+5a+4)+(aa+3a+2)+(aa+2a))
a(a+2)(a+1)(a+4)

Combine like terms in the numerator.

3aa+10a+6
a(a+2)(a+1)(a+4)

Well, I'm stumped. Lets multiply the lower half and see if there's a common factor.

3aa+10a+6
(aa+2a)(aa+5a+4)

3aa+10a+6
a^4+7a^3+14aa+8a

Looks like a dead end to me, and if we factor the numerator we get (a+(5+sqrt(7))/3)(a+(5-sqrt(7))/3).

Sooo, it seems it's time to do the mathematical thing and generalize in a random direction!

In 1/15+1/21+1/28 you have 3 terms with equal numerators, and the harmonic mean of the denominators is 20. The answer is 3/20. This obviously works for anything with only 1 term, and if you think of the process of taking the harmonic mean, you see that you're only doing the same steps more elaborately, so you might as well just add the fractions you started with....
1/17+1/3 harmonic mean=51/10
1/17+1/3=20/51.

Wow, I've done nothing.

If you feel like it, have your calculator graph

3XX+10X+6
X^4+7X^3+14XX+8X

or

1/(X(X+2)) + 1/(X(X+4)) + 1/((X+1)(X+4))

You'll see that they're equivalent, and quite without shortcuts. Just add the fractions, lazy!

Guys:

I have to wonder whether future problems will fit the type solved above, namely that the factors are specificly of the type a, a+1, a+2, a+4.

I have found a solution for a more general form of the problem, and would like to ask someone to give it a try to see what they can come up with on their own for the following form:

1/ab + 1/ac + 1/cd, where a is a common factor between denominators 1 & 2, c is a common factor between denominators 2 & 3, and b and d are the factors not common among any of the 3 denominators. You should find that the problem written in this form (as well as the solution you come up with) will work for the specific problem listed above.

Steve

Vinay:

Yes and I prefer to show it as (AB + CD + bd)/[(AB)(CD)]. Now if you remember the problem will appear as

1/ab + 1/ac + 1/cd. So all that is required to sum the numerator is to add the following: the 1st fraction numerator (ab), the 3rd fraction numerator (cd) and the product of the two factors that do not appear in multiple denominators (b*d). Then to obtain the denominator simply multiply the first and third denominators together. So for your problem

1/15 + 1/21 + 1/28

the solution is simply (15+28+5*4)/[(15)(28)] or 63/[3*5*4*7] which reduces to 3/20. To me this is a relatively easy formula becuase I can look directly on the paper for 4 of the 5 terms in the formula (those shown in capital letters above).