I want to find the sum $$S=f\left(\dfrac{1}{2012} \right) +f\left(\dfrac{2}{2012} \right) +\cdots + f\left(\dfrac{2011}{2012} \right), $$ where $$f(x) = \dfrac{4^x}{4^x + 2}.$$ I tried
f[x_] := 4^x/(4^x + 2)
Sum[f[i/2012], {i, 1, 2011}]
But I can't get the answer. How do I tell Mathematica to do that?
I know, $$f(x) + f(1 - x) =1.$$
I used
Simplify[f[x] + f[1 - x]]
The Simplify idea is on the right track, but you have to tell Mathematica about the relationship between f and Plus in a slightly different way:
ClearAll[f]; Plus[f[x_], f[y_]] /; x + y == 1 ^:= 1
The Upset symbol ^:= means this is a definition for f rather than Plus. This approach is inefficient--there is going to be a lot of blind checking to find pairs of arguments x and y that sum to unity--but it works.
With[{n = 2012}, Sum[f[i/n], {i, 1, n - 1}]]
$1005+f\left[\frac{1}{2}\right]$
Ah, yes: we do have to remember to give a definition of f in order to handle any leftover terms from the sum that do not simplify:
Block[{x}, % /. f[x_] -> 4^x/(4^x + 2)]
$\frac{2011}{2}$
Despite the inefficiency, this approach clearly and suggestively captures the point of the problem: to use that sum-to-unity identity for $f$ to simplify the sum, leaving only the middle term $f(\frac{1}{2})$ to be evaluated.
I like FindSequenceFunction it can be pretty awesome.
Here I give it the sequence (your case is when k=2012):
$$ \left\{ \sum_{n=1}^{k-1} f \left( \frac{i}{n} \right) \right\}_{k=1}^{25} $$
partial=Rationalize@NSum[f[i/#],{i,1,#-1}]&/@Range[1,25]
g=FindSequenceFunction[partial]
(* 1/2 (-1+#1)& *)
g[10000] == Rationalize@NSum[f[i/10000],{i,1,9999}]
(* True *)
Bonus toy rule replacement that works for small numbers
n=31;
Sum[ff[i/n],{i,1,n-1}]//.Plus[ff[i_],ff[j_]]/;(i==1-j):>1
(* 15 *)
n=30;
Sum[ff[i/n],{i,1,n-1}]//.Plus[ff[i_],ff[j_]]/;(i==1-j):>1
(* 14+ff[1/2] *)
%/.ff->f
(* 29/2 *)
g/@{31,30}
(* {15,29/2} *)
The neat thing about this is that it hints of an obvious extension to other functions as long as they satisfy f[x]+f[1-x]==1
You could also use the knowledge that $f(x) + f(1-x) = 1$ as follows:
f[x_] := 4^x/(4^x + 2)
Sum[Simplify[f[i/2012] + f[(2012 - i)/2012]], {i, 1, 2011}]/2
$\frac{2011}{2}$
Here I just did the sum twice - once forward and once backward. Then I divided by two. The assumption about $f$ actually never has to be explained to Mathematica here, I just helped it along by arranging the terms so that Simplify can recognize them inside the sum.
NSum[f[i/2012], {i, 1, 2011}]$\endgroup$