(1)求数列1,1/1+2 ,1/1+2+3 ……1/1+2+3+n的通项公式Αn (2)求数列{(1)求数列1,1/1+2 ,1/1+2+3 ……1/1+2+3+n的通项公式Αn(2)求数列{An}和前n项和Sn
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![(1)求数列1,1/1+2 ,1/1+2+3 ……1/1+2+3+n的通项公式Αn (2)求数列{(1)求数列1,1/1+2 ,1/1+2+3 ……1/1+2+3+n的通项公式Αn(2)求数列{An}和前n项和Sn](/uploads/image/z/10358309-29-9.jpg?t=%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%971%2C1%2F1%2B2+%2C1%2F1%2B2%2B3+%E2%80%A6%E2%80%A61%2F1%2B2%2B3%2Bn%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%CE%91n+%EF%BC%882%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9B%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%971%2C1%2F1%2B2+%2C1%2F1%2B2%2B3+%E2%80%A6%E2%80%A61%2F1%2B2%2B3%2Bn%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%CE%91n%EF%BC%882%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9BAn%EF%BD%9D%E5%92%8C%E5%89%8Dn%E9%A1%B9%E5%92%8CSn)
(1)求数列1,1/1+2 ,1/1+2+3 ……1/1+2+3+n的通项公式Αn (2)求数列{(1)求数列1,1/1+2 ,1/1+2+3 ……1/1+2+3+n的通项公式Αn(2)求数列{An}和前n项和Sn
(1)求数列1,1/1+2 ,1/1+2+3 ……1/1+2+3+n的通项公式Αn (2)求数列{
(1)求数列1,1/1+2 ,1/1+2+3 ……1/1+2+3+n的通项公式Αn
(2)求数列{An}和前n项和Sn
(1)求数列1,1/1+2 ,1/1+2+3 ……1/1+2+3+n的通项公式Αn (2)求数列{(1)求数列1,1/1+2 ,1/1+2+3 ……1/1+2+3+n的通项公式Αn(2)求数列{An}和前n项和Sn
(1)An =2/[n(n+1)]
分母是前n个正整数的和,是0.5n(n+1)
(2)Sn=1+1/3+1/6+……+2/[n(n+1)]
=2{1/2+1/6+1/12+……+1/[n(n+1)]}
=2{1/(1×2)+1/(2×3)+1/(3×4)+……+1/[n(n+1)]}
=2{(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+[1...
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(1)An =2/[n(n+1)]
分母是前n个正整数的和,是0.5n(n+1)
(2)Sn=1+1/3+1/6+……+2/[n(n+1)]
=2{1/2+1/6+1/12+……+1/[n(n+1)]}
=2{1/(1×2)+1/(2×3)+1/(3×4)+……+1/[n(n+1)]}
=2{(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+[1/n-1/(n+1)]}
=2[1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)
收起
1.
An=2/[n*(n+1)]
2.
Sn=2*[1/(1×2)+1/(2×3)+...+1/n*(n+1)]
=2*[1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=2*[1-1/(n+1)]
=2n/(n+1)