bn=a1+a2+a3+.......+an−1+an olsun
mi ve ki dizi olsunlar;
mi=m1,m2,m3,m4,.........mn−1,mn
ki=k1,k2,k3,k4,.........kn−1,kn
Amacım genel gösterimi vermek , birazda olsa ki ve mi ler hakkında fıkır zemini oluşturmak.
Tek tek bazı açılımlar yapalımki genel gösterime tam uysunlar....
ana kural
(x+y)n=n∑k=0(nk).xk.yn−k
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bmnn=[bn−1+an]mn=mn∑kn=0(mnkn)bkn(n−1)a(mn−kn)n
bkn(n−1)=[bn−2+a(n−1)]kn=kn∑mn−1=0(knmn−1)bm(n−1)(n−2)a(kn−m(n−1))(n−1)
bmn−1n−2=[bn−3+an−2]mn−1=mn−1∑kn−1=0(mn−1kn−1)bkn−1(n−3)a(mn−1−kn−1)n−2
bkn−1(n−3)=[bn−4+a(n−3)]kn−1=kn−1∑mn−2=0(kn−1mn−2)bm(n−2)(n−4)a(kn−1−m(n−2))(n−3)
bmn−2n−4=[bn−5+an−4]mn−2=mn−2∑kn−2=0(mn−2kn−2)bkn−2(n−5)a(mn−2−kn−2)n−4
bkn−2(n−5)=[bn−6+a(n−5)]kn−2=kn−2∑mn−3=0(kn−2mn−3)bm(n−3)(n−6)a(kn−2−m(n−3))(n−5)
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yani
Her birini bir öncekinde yerine koyarsak.....
bmnn=[bn−1+an]mn=mn∑kn=0(mnkn)[kn∑mn−1=0(knmn−1)[mn−1∑kn−1=0(mn−1kn−1)[kn−1∑mn−2=0(kn−1mn−2)[⋱⋱⋱⋱]a(kn−1−m(n−2))(n−3)]a(mn−1−kn−1)n−2]a(kn−m(n−1))(n−1)]a(mn−kn)n