作业帮设各项均为正数的数列an的前n项和为Sn,且Sn满足Sn²-(n²+n-3)Sn-3(n²+n)=0(1)求a1的值；(2)求数列｛an｝的通项公式；(3)求1/[a1(a1+2)] +1/[a2(a2+2)]+…+1/[an(an+2)].
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![设各项均为正数的数列an的前n项和为Sn,且Sn满足Sn²-(n²+n-3)Sn-3(n²+n)=0(1)求a1的值；(2)求数列｛an｝的通项公式；(3)求1/[a1(a1+2)] +1/[a2(a2+2)]+…+1/[an(an+2)].](/uploads/image/z/8663911-7-1.jpg?t=%E8%AE%BE%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B0%E7%9A%84%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94Sn%E6%BB%A1%E8%B6%B3Sn%26%23178%3B-%28n%26%23178%3B%2Bn-3%29Sn-3%28n%26%23178%3B%2Bn%29%3D0%281%29%E6%B1%82a1%E7%9A%84%E5%80%BC%EF%BC%9B%282%29%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B%283%29%E6%B1%821%2F%5Ba1%28a1%2B2%29%5D+%2B1%2F%5Ba2%28a2%2B2%29%5D%2B%E2%80%A6%2B1%2F%5Ban%28an%2B2%29%5D.)
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设各项均为正数的数列an的前n项和为Sn,且Sn满足Sn²-(n²+n-3)Sn-3(n²+n)=0(1)求a1的值；(2)求数列｛an｝的通项公式；(3)求1/[a1(a1+2)] +1/[a2(a2+2)]+…+1/[an(an+2)].
设各项均为正数的数列an的前n项和为Sn,且Sn满足Sn²-(n²+n-3)Sn-3(n²+n)=0
(1)求a1的值；(2)求数列｛an｝的通项公式；(3)求1/[a1(a1+2)] +1/[a2(a2+2)]+…+1/[an(an+2)].
设各项均为正数的数列an的前n项和为Sn,且Sn满足Sn²-(n²+n-3)Sn-3(n²+n)=0(1)求a1的值；(2)求数列｛an｝的通项公式；(3)求1/[a1(a1+2)] +1/[a2(a2+2)]+…+1/[an(an+2)].
（1）因为Sn²-(n²+n-3)Sn-3(n²+n)=0
所以(Sn+3)[Sn-(n²+n)]=0
因为数列an的各项均为正数
所以Sn-(n²+n)=0
Sn=n²+n
a1=S1=2
（2）an=Sn-S(n-1)=2n(n≥2)
当n=1时,a1=2×1=2,所以a1符合通项公式
故数列｛an｝的通项公式为an=2n
（3）1/[a1(a1+2)] +1/[a2(a2+2)]+…+1/[an(an+2)]
=1/(2×4) +1/(4×6)+…+1/[an(an+2)]
=1/2×[1/2-1/4+1/4-1/6+…+1/an-1/(an+2)]
=1/2×[1/2-1/(an+2)]
=n/(4n+4)