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2

Update exp = Subscript[S, 3] Subscript[S, 4] + Subscript[S, 2] Subscript[S, 3] + Subscript[S, 4] Subscript[S, 5] $S_2 S_3+S_4 S_3+S_4 S_5$ This code toSum check for able form of Sum and if possible then it is applied Defer. toSum[exp_] := Module[{l = List @@ exp, l2, s}, l = l /. Subscript[S, a_] Subscript[S, b_] -> {a, b}; s = l[[1, 1]]; ...


2

f[n_] := Total[Total[#] - Min[#] & /@ Tuples[Range[6], n]]/6^n;


2

f[n_] := 1/6^n Module[{xx = Array[x, n]}, Sum[Tr@xx - Min @@ xx, Evaluate[Sequence @@ ({#, 6} & /@ xx)]]]


4

You can use the Partition: data= {16, 13, 42, 26, 35, 39, 47, 49, 1, 9}; Times @@@ Partition[data, 2, 1] {208, 546, 1092, 910, 1365, 1833, 2303, 49, 9} The summation of the list: Plus@@(Times @@@ Partition[data, 2, 1]) 8315


5

You can use MovingMap in version 10. data = RandomInteger[{1, 50}, 10] (* {16, 13, 42, 26, 35, 39, 47, 49, 1, 9} *) MovingMap[Times @@ # &, data, {2, Left}] (* {208, 546, 1092, 910, 1365, 1833, 2303, 49, 9} *) There you can see the items muliplied by the pairs. Now just Apply Plus to get the total. Plus @@ MovingMap[Times @@ # &, data, {2, ...


4

In order to get the expression yo simplify to the fullest extent, one could make the additional assumption that the label $\alpha$ always falls within the range of the summation whose index is $\beta$. To do this, we have to add some rules which are easier to write if the starting expression is brought into a slightly different form: $$\frac{\partial ...


2

It is too difficult for Mathematica to combine two sums. Even in the following simple example 2 Sum[a[n], {n, 1, q}] - Sum[2 a[n], {n, 1, q}] (* 2 Sum[a[n], {n, 1, q}] - Sum[2 a[n], {n, 1, q}] *) There is only one exception: if your expressions are exactly the same (not necessarily Sum) they will subtracted to 0 Sum[a[n], {n, 1, q}] - Sum[a[n], {n, 1, ...


-2

I can't get it to work with "q", but I can get it to work with the dummy variables m & n. Sum[f[n], {n, 1, 10}] - Sum[f[m], {m, 1, 10}] // FullSimplify 0 As you can see, I replaced your subscript with functional notation.


3

These sums are conditionally convergent, so you have to specify a summation order that suits your purpose. See Wikipedia. Since the question seems to be focused on the issue of dropping a term in a sum, here is one way of doing it, without making any claim that the resulting sum is of any use: Chop[NSum[ If[i == j == k == 0, 0, (-1)^(i + j + ...



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