前>>153
>>161
k=4のとき1/(k^4-10k^2+9)=1/(256-160+9)
=1/105
=1/1･3･5･7
={(1/1-1/7)(1/6)-(1/3-1/5)(1/2)}(1/8)
k=5のとき1/(k^4-10k^2+9)=1/(625-250+9)
=1/384
=1/2･4･6･8
={(1/2-1/8)(1/6)-(1/4-1/6)(1/2)}(1/8)

k=nのとき1/(n^4-10n^2+9)=1/(n^2-1)(n^2-9)
=1/(n-3)(n-1)(n+1)(n+3)
=[{1/(n-3)-1/(n+3)}(1/6)-{1/(n-1)-1/(n+1)}(1/2)](1/8)
与式=Σ[k=4,n] 1/(k^4-10k^2+9)
={(1/1-1/7)(1/6)-(1/3-1/5)(1/2)}(1/8)+
{(1/2-1/8)(1/6)-(1/4-1/6)(1/2)}(1/8)+
{(1/3-1/9)(1/6)-(1/5-1/7)(1/2)}(1/8)+
{(1/4-1/10)(1/6)-(1/6-1/8)(1/2)}(1/8)+……+
[{1/(n-3)-1/(n+3)}(1/6)-{1/(n-1)-1/(n+1)}(1/2)](1/8)
±0になって相殺する法則がみつかればもっと簡単になるはず。とりあえず48で通分か。