09/03/28 01:47:25
>>270
1/[(x+1)(x+2)(x+3)] = (1/2){1/[(x+1)(x+2)] - 1/[(x+2)(x+3)]},
1/[(x+1)(x+2)] = 1/(x+1) - 1/(x+2),
1/[(x+2)(x+3)] = 1/(x+2) - 1/(x+3),
を使う。
1/{x(x+1)(x+2)(x+3)･････(x+n)} = (1/n){1/[x(x+1)(x+2)･･･(x+n-1)] - 1/[(x+1)(x+2)(x+3)･･････(x+n)]},