1、
最后应该是1/(96×98×100)吧?不然前面分母都是偶数,最后一项是分母奇数,没有规律可循。
考察一般项:
1/[(2n)(2n+2)(2n+4)]=(1/8)[1/[n(n+1)(n+2)]]=(1/16)[(1/n -1/(n+1)-(1/(n+1)-1/(n+2)]
1/(2×4×6)+1/(4×6×8)+...+1/(96×98×100)
=(1/16)[(1/1-1/2)-(1/2-1/3)+(1/2-1/3)+(1/3-1/4)+...+(1/48-1/49)-(1/49-1/50)]
=(1/16)[(1/1-1/2)-(1/49-1/50)]
=(1/16)(1-1/2 -1/49+1/50)
=1237/39600
2、
考察一般项:
1/[(2n-1)(2n+1)(2n+3)]=(1/8)[[1/(2n-1)-1/(2n+1)]-[1/(2n+1)-1/(2n+3)]]
1/(1×3×5)-1/(3×5×7)-...-1/(95×97×99)
=2/(1×3×5)-[1/(1×3×5)+1/(3×5×7)+...+1/(95×97×99)]
=2/15 -(1/8)[(1/1-1/3)-(1/3-1/5)+(1/3-1/5)-(1/5-1/7)+...+(1/95-1/97)-(1/97-1/99)]
=2/15 -(1/8)(1-1/3 -1/97+1/99)
=2/15 -800/9603
=(6402-4000)/48015
=2402/48015
3、
最后应该是840吧。修改后再做,方法和上面的一样的。