数列an的前n项和为Sn.且满足a1=1.2Sn=(n+1)an

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数列an的前n项和为Sn.且满足a1=1.2Sn=(n+1)an
（1）求an与a（n-1）的关系式.并求an的通项公式
（2）求和Wn=1/（a2^2-1）+1/（a3^2-1）+……+1/（a（n+1）^2-1）

2·a(n) = 2[Sn - S(n-1)] = (n+1)an - n·a(n-1)
∴(n-1)an = n·a(n-1) ,∴an/[a(n-1)] = n/(n-1) ,.,a3/a2 = 3/2 ,a2/a1 = 2/1 ,将上述式子相乘：an/a1 = n ,∴an = n·1 = n ,
即：{an}是以1为首项、1为公差的等差数列 .
1/(a2^2-1) = 1/(2^2-1) = 1/2 - (1/6)
1/(a3^2-1) = 1/4 - (1/8)
1/(a4^2-1) = 1/6 - (1/10)
1/(a5^2-1) = 1/8 - (1/12)
.
1/{[a(n)]^2-1} = 1/{2[a(n) - 1]} - 1/{2[a(n) + 1]]}
= 1/{2[a(n) - 1]} - 1/[2a(n+1)]
1/{[a(n+1)]^2-1} = 1/{2[a(n+1) - 1]} - 1/{2[a(n+1) + 1]]}
= 1/[2a(n)] - 1/{2[a(n+1) + 1]]}
∴Wn = (1/2)·[1 + (1/2) - (1/n+1) - (1/n+2)]
= (3/4) - [1/(2n+2)] - [1/(2n+4)]

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