■ある無限級数(その104)
(その103)を有限級数版にすると・・・
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S=1+2x+3x^2+・・・+nx^n-1
(1−x)S=1+x+x^2+・・・+x^n-1−nx^n
=(1−x^n)/(1−x)−nx^n
S={1−(n+1)x^n+nx^n+1}/(1−x)^2
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T=1+4x+9x^2+・・・+n^2x^n-1
(1−x)T=1+3x+5x^2+・・・+(2n−1)x^n-1−n^2x^n
=2S−(1+x+x^2+・・・+x^n-1)−n^2x^n
T={1+x−(n+1)^2x^n+(2n^2+2n−1)x^n+1−n^2x^n+2}/(1−x)^3
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