■カタラン数と漸化式(その11)
(n+1)^k−(n+1)=(k,1)Sk-1+(k,2)Sk-2+・・・+(k,k−1)S1
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[1]k=2を代入
(n+1)^2−(n+1)=2S1
S1=n(n+1)/2
[2]k=3を代入
(n+1)^3−(n+1)=3S2+3S1
n(n^2+3n+2)=3S2+3S1
S2={2n(n+1)(n+2)−3n(n+1)}/6
S2=n(n+1)(2n+1)/6
[3]k=4を代入
(n+1)^4−(n+1)=4S3+6S2+4S1
n(n^3+4n^2+6n+3)=4S3+n(n+1)(2n+1)+2n(n+1)
n(n+1)(n^2+3n+3)=4S3+n(n+1)(2n+1)+2n(n+1)
n(n+1)(n^2+3n+3−2n−1−2)=4S3
n^2(n+1)^2/4=S3
S1=Σk=n(n+1)/2
S2=Σk^2=n(n+1)(2n+1)/6
S3=Σk^3=n^2(n+1)^2/4
は多くの読者にとってお馴染みの公式であろう.
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さらに,
S4=Σk^4=n(n+1)(2n+1)(3n^2+3n−1)/30
S5=Σk^5=n^2(n+1)^2(2n^2+2n−1)/12
S6=Σk^6=n(n+1)(2n+1)(3n^4+6n^3−3n+1)/42
S7=Σk^7=n^2(n+1)^2(3n^4+6n^3−n^2−4n+2)/24
S8=Σk^8=n(n+1)(2n+1)(5n^6+15n^5+5n^4−15n^3−n^2+9n−3)/90
と続く.
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