急~~~高一数学数列题~

1,
1/(1*3) + 1/(2*4) + 1/(3*5) + ...+ 1/[(n-2)n] + 1/[(n-1)(n+1)] + 1/[n(n+2)]
= (1/2)[1/1 - 1/3 + 1/2 - 1/4 + 1/3 - 1/5 + ... + 1/(n-2) - 1/n + 1/(n-1) - 1/(n+1) + 1/n - 1/(n+2)]
= (1/2)[1/1 + 1/2 - 1/(n+1) - 1/(n+2)]
= (1/2){3/2 - (2n+3)/[(n+1)(n+2)]}
= 3/4 - (2n+3)/[2(n+1)(n+2)]

2,
1/[1*2] + 1/[2*3] + ... + 1/[n*(n+1)]
= 1/1 - 1/2 + 1/2 - 1/3 + ... + 1/n - 1/(n+1)
= 1 - 1/(n+1)
= n/(n+1)

3,
a(n) = 1/[n^(1/2) + (n+1)^(1/2)]
= [(n+1)^(1/2) - n^(1/2)]

S(n) = 2^(1/2) - 1^(1/2) + 3^(1/2) - 2^(1/2) + ... + (n+1)^(1/2) - n^(1/2)
= (n+1)^(1/2) - 1

1题,1/n(n+2)=1/2*(1/n-1/(n+2))
原式*2=1-1/3+1/2-1/4+1/3-1/5+1/4-1/6.....+1/n-1/(n+2)
中间可消掉，=1+1/2-1/(n+1)-1/(n+2),结果再除以2.
2题，1/n(n+1)=1/n-1/(n+1),方法和上题近似。
3题，分子分母同乘根号（n+1）-根号n
an=根号（n+1）-根号n。相加可抵消