■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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