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\(\displaystyle{ per(a,b)^{n}=\frac{(a+b)^{n}+a^{n}+b^{n}}{2}}\)

\(\displaystyle{ per(a,b,c)^{n}=\frac{(a+b+c)^{n}+a^{n}+b^{n}+c^{n}}{2}}\)

\(\displaystyle{ per(a,b,c,d)^{n}=\frac{(a+b+c+d)^{n}+a^{n}+b^{n}+c^{n}+d^{n}}{2}}\)

To najnowszy wzór, na permutację, i właściwie najlepszy jaki się trafił.

Tu możesz wstawić dowolną, liczbę pod indeks.:



\(\displaystyle{ \frac{95x^{5}+2x^{4}+0x^{3}+0x^{2}+0x+0}{(x+1)(x+2)(x-1)(x-2)}}\)


\(\displaystyle{ 95x+}\)

\(\displaystyle{ -95 per(1,2,-1-,2)^{1}+2+}\)

\(\displaystyle{ \frac{-95 per(1,2,-1-,2)^{2}+2per(1,2,-1-,2)^{1}-0}{(x+1)}}\)

\(\displaystyle{ \frac{+95 per(1,2,-1,-2)^{3}-2per(1,2,-1,-2)^{2}+0per(1,2,-1,-2)^{1}-0}{(x+1)(x+2)}}\)

\(\displaystyle{ \frac{+95 per(1,2,-1,-2)^{4}-2per(1,2,-1,-2)^{3}+0per(1,2,-1,-2)^{2}-0per(1,2,-1,-2)^{1}+0}{(x+1)(x+2)(x-1)}}\)

\(\displaystyle{ \frac{-95(-2)^{5}+2(-2)^{4}-0(-2)^{3}+0(-2)^{2}-0(-2)+0(-2)}{(x+1)(x+2)(x-1)(x-2)}}\)


\(\displaystyle{ per(1,2,-1-2)^{1}=0}\)

\(\displaystyle{ per(1,2,-1,-2)^{2}=\frac{(1+2+-1+-2)^{2}+(1)^{2}+(2)^{2}+(-1)^{2}+(-2)^{2}}{2}= 5}\)

\(\displaystyle{ per(1,2,-1,-2)^{3}=\frac{(1+2+-1+-2)^{3}+(1)^{3}+(2)^{3}+(-1)^{3}+(-2)^{3}}{2}= 0}\)

\(\displaystyle{ per(1,2,-1,-2)^{4}=\frac{(1+2+-1+-2)^{4}+(1)^{4}+(2)^{4}+(-1)^{4}+(-2)^{4}}{2}= 17}\)


\(\displaystyle{ 95x+}\)

\(\displaystyle{ 2+}\)

\(\displaystyle{ \frac{-95\cdot5}{(x+1)}+}\)

\(\displaystyle{ \frac{-2\cdot 5}{(x+1)(x+2)}+}\)

\(\displaystyle{ \frac{+95\cdot 17}{(x+1)(x+2)(x-1)}+}\)

\(\displaystyle{ \frac{-95\cdot (-32)+2\cdot16}{(x+1)(x+2)(x-1)(x-2)}}\)

Dodano po 1 godzinie 8 minutach 43 sekundach:
Przez te lata, odkryłem kilkadziesiąt, wzorów, na sumę permutacji. Ten jest najlepszy.

Dodano po 3 minutach 39 sekundach:
Ten też, jest niezły:

\(\displaystyle{ per(a,b,c)^{1}=(a+b+c)}\)
\(\displaystyle{ per(a,b,c)^{2}=(a(a+b+c)+b(b+c)+cc}\)
\(\displaystyle{ per(a,b,c)^{3}=((a+b+c)(a(a+b+c)+b(b+c)+cc)}\)
\(\displaystyle{ per(a,b,c)^{4}=((a+b+c)^{2}(a(a+b+c)+b(b+c)+cc)}\)
\(\displaystyle{ per(a,b,c)^{5}=((a+b+c)^{2}(a(a+b+c)^{2}+b(b+c)^{2}+cc^{2})}\)
\(\displaystyle{ per(a,b,c)^{6}=((a+b+c)^{3}(a(a+b+c)^{2}+b(b+c)^{2}+cc^{2})}\)
\(\displaystyle{ per(a,b,c)^{7}=((a+b+c)^{4}(a(a+b+c)^{2}+b(b+c)^{2}+cc^{2})}\)
\(\displaystyle{ per(a,b,c)^{8}=((a^{2}+b^{2}+c^{2})(a+b+c)^{3}(a(a+b+c)^{2}+b(b+c)^{2}+cc^{2})}\)
\(\displaystyle{ per(a,b,c)^{9}=((a^{2}+b^{2}+c^{2})(a+b+c)^{4}(a(a+b+c)^{2}+b(b+c)^{2}+cc^{2})}\)
\(\displaystyle{ per(a,b,c)^{10}=((a^{2}+b^{2}+c^{2})(a+b+c)^{5}(a(a+b+c)^{2}+b(b+c)^{2}+cc^{2})}\)
\(\displaystyle{ per(a,b,c)^{11}=((a^{2}+b^{2}+c^{2})^{2}(a+b+c)^{4}(a(a+b+c)^{2}+b(b+c)^{2}+cc^{2})}\)

Dodano po 14 minutach 18 sekundach:
Ten jest fajny, ale działa dopiero od piątej permutacji.

\(\displaystyle{ per(a,b,c)^{8}=(a+b+c)(a^{2}+b^{2}+c^{2}) \cdot per(a,b,c)^{5}}\)
\(\displaystyle{ per(a,b,c)^{9}=(a+b+c)(a^{2}+b^{2}+c^{2}) \cdot per(a,b,c)^{6}}\)
\(\displaystyle{ per(a,b,c)^{10}=(a+b+c)(a^{2}+b^{2}+c^{2}) \cdot per(a,b,c)^{7}}\)
\(\displaystyle{ per(a,b,c)^{11}=(a+b+c)(a^{2}+b^{2}+c^{2}) \cdot per(a,b,c)^{8}}\)
\(\displaystyle{ per(a,b,c)^{12}=(a+b+c)(a^{2}+b^{2}+c^{2}) \cdot per(a,b,c)^{9}}\)
\(\displaystyle{ per(a,b,c)^{13}=(a+b+c)(a^{2}+b^{2}+c^{2}) \cdot per(a,b,c)^{10}}\)
\(\displaystyle{ per(a,b,c)^{14}=(a+b+c)(a^{2}+b^{2}+c^{2}) \cdot per(a,b,c)^{11}}\)
\(\displaystyle{ per(a,b,c)^{15}=(a+b+c)(a^{2}+b^{2}+c^{2}) \cdot per(a,b,c)^{12}}\)