I know
- S[n] is Sum[a[k], {k,1,n}]
- a[1]==1
I want to get the general solution of a[n] and S[n].
So I tried this one.
But I can’t get the answer. What can I do for this problem?
I do not understand what S_k...(*) means. Maybe this approach is helpful:
s[n_] := Sum[a[i], {i, 1, n}]
ss[n_] := Sum[(n - i + 1) a[i], {i, 1, n}]
(* your equation: *)
eqn[n_] := (n + 1) s[n + 1] == Log[2, n + 2] + ss[n]
nMax = 10; (* as an example *)
vars = Table[a[n], {n, 1, nMax + 1}];
equations = Table[eqn[n], {n, 0, nMax}];
(* solutions: *)
Solve[equations, vars]
Though not rigorous according to this it seems that
a[n]= -\frac{\log(n) - \log(n+1)}{n \log(2)}
is the general solution.
Oh. Thank you very much. Your approach is helpful for me.
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Looking at your equations eqn[ n ] (in Henriks notation). expanding them (by hand ) and and forming the difference eqn[ j + 1 ] - eqn [ j ] you arrive at
( j + 2 ) a[ j + 2 ] == Log[ ( j + 3 ) / ( j + 2 ) ] / Log[ 2 ]
and therefore
a[ j_] := Log[(1 + j)/j]/(j Log[2])