■カタラン数と漸化式(その6)
Sk=1^k+2^k+3^k+・・・+n^k
とおく.
(x+1)^k=x^k+(k,1)x^k-1+(k,2)x^k-2+・・・+(k,k−1)x+1
(x+1)^k−x^k=(k,1)x^k-1+(k,2)x^k-2+・・・+(k,k−1)x+1
x=1,2,3,・・・,nとして,両辺を加えると
(n+1)^k−1=(k,1)Sk-1+(k,2)Sk-2+・・・+(k,k−1)S1+n
(n+1)^k−(n+)=(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
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