数列{an}中,a1=1/2,a(n+1)=(nan)/[(n+1)*(nan+1)],其前n项的和为S
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数列{an}中,a1=1/2,a(n+1)=(nan)/[(n+1)*(nan+1)],其前n项的和为S
(1)设bn=1/an,求证{bn}是等差数列
(2) 求Sn的表达式
(1)设bn=1/an,求证{bn}是等差数列
(2) 求Sn的表达式
a(n+1)=(nan)/[(n+1)*(nan+1)]
两边乘以n+1,在取倒数
1/[a(n+1)*(n+1)] = 1 + 1/(nan)
所以b(n+1) = 1 + bn ,即bn是等差数列(lz你应该是题目输错了 bn=1/nan )
(2)、由于bn是等差数列 bn = b1 + 1 * (n-1)
= 2 + n-1
= n+1
bn = 1/(nan) => an = 1/[n*(n+1)]
= 1/n - 1/(n+1)
Sn = 1 - 1/2 + 1/2 - 1/3 + …… + 1/n -1/(n+1)
= 1- 1/(n+1)
= n/(n+1)
两边乘以n+1,在取倒数
1/[a(n+1)*(n+1)] = 1 + 1/(nan)
所以b(n+1) = 1 + bn ,即bn是等差数列(lz你应该是题目输错了 bn=1/nan )
(2)、由于bn是等差数列 bn = b1 + 1 * (n-1)
= 2 + n-1
= n+1
bn = 1/(nan) => an = 1/[n*(n+1)]
= 1/n - 1/(n+1)
Sn = 1 - 1/2 + 1/2 - 1/3 + …… + 1/n -1/(n+1)
= 1- 1/(n+1)
= n/(n+1)
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