Given the goal, I wonder if the following approach might be of use. You are operating on explicit Sum objects. Perhaps commute this, that is, do the operation(s) on the general form of the jth term first, and then Sum over j.

I'm a statistician and use combinatorics in a lot of the work I do. (My major professor who was a statistician co-wrote a book called Combinatorial Chance - I highly recommend looking at that book for some of the things you might want to do with such sums as you've described.)

I also highly recommend the book called Mathematical Statistics with Mathematica written by Colin Rose and Murray Smith (which describes what's in the mathStatica package I mentioned earlier). Chapter 7 deals with augmented symmetric functions.

About specifying n....To figure out general solutions using sums like you've used above sometimes one needs to specify a suite of particular n's and hopefully a pattern will be recognized. Or you try 1, 2, 3, 4, ... and some recursive pattern becomes apparent. But with augmented symmetric functions you won't need to specify n to obtain the general results.

It looks like you might be trying to re-invent power sums and symmetric functions used in lots of combinatorial problems (which is a good thing). The power sums and symmetric polynomials are found in Mathematica and the add-on package mathStatica.

Is the following something that you're trying to do?

a = {a1, a2, a3, a4, a5};
    b = {b1, b2, b3, b4, b5};
    n = 5;
    FullSimplify[Sum[(a[[i]] + b[[i]]) (a[[i]] + b[[i]]), {i,n}]/(Sum[a[[i]]^2, {i, n}] + 2*Sum[a[[i]] b[[i]], {i, n}] + Sum[b[[i]]^2, {i, n}])]

Hi Bruce,

You are right. Thank you very much for helping me with subscripts.

Rewriting my question, this is now my issue. The following, which I believe to be equal to one, does not simplify

FullSimplify[
  Sum[(Subscript[a, i] + Subscript[b, i])*(Subscript[a, i] + Subscript[b, i]), i] /
      (Sum[Subscript[a, i]*Subscript[a, i], i] + 2 * Sum[Subscript[a, i]*Subscript[b, i], i] + Sum[Subscript[b, i]*Subscript[b, i], i])
]

instead it just gives back the pretty printed version of the equation. Interestingly, I was able to use the Blank[] notation to get the behavior that I wanted, so I know Mathematica knows how to do this.

FullSimplify[
  Sum[(a_i + b_i)*(a_i + b_i), {i, 1}]/(Sum[a_i*a_i, {i, 1}] + 2 * Sum[a_i*b_i, {i, 1}] + Sum[b_i*b_i, {i, 1}])
]

returns 1 and does some other algebra that I've tested as I expect, but it does make extremely verbose output.

I should mention that I'm not just interested in this type of sum but also sums of the form Sum[ai * bi, i] / Sum[ai * ai, i] where removing the _i would give an incorrect answer.