{an}中,a1=1/2,an+1=nan/(n+1)(nan+1),n∈正整数,设bn=1/nan,求证(1){bn}是等差数列(2)Sn的表达式
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![{an}中,a1=1/2,an+1=nan/(n+1)(nan+1),n∈正整数,设bn=1/nan,求证(1){bn}是等差数列(2)Sn的表达式](/uploads/image/z/13893600-48-0.jpg?t=%7Ban%7D%E4%B8%AD%2Ca1%3D1%2F2%2Can%2B1%3Dnan%2F%28n%2B1%29%28nan%2B1%29%2Cn%E2%88%88%E6%AD%A3%E6%95%B4%E6%95%B0%2C%E8%AE%BEbn%3D1%2Fnan%2C%E6%B1%82%E8%AF%81%281%29%7Bbn%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%282%29Sn%E7%9A%84%E8%A1%A8%E8%BE%BE%E5%BC%8F)
{an}中,a1=1/2,an+1=nan/(n+1)(nan+1),n∈正整数,设bn=1/nan,求证(1){bn}是等差数列(2)Sn的表达式
{an}中,a1=1/2,an+1=nan/(n+1)(nan+1),n∈正整数,设bn=1/nan,求证(1){bn}是等差数列(2)Sn的表达式
{an}中,a1=1/2,an+1=nan/(n+1)(nan+1),n∈正整数,设bn=1/nan,求证(1){bn}是等差数列(2)Sn的表达式
(1)a(n+1)=nan/(n+1)(nan+1),
移项,(n+1)a(n+1)=nan/(nan+1)
两边取倒数,1/(n+1)a(n+1)=1+1/nan
bn=1/nan,所以b(n+1)-bn=1,b1=1/(1/2)=2
即bn=1+n,为等差数列
(2)
an=1/n(n+1)=1/n -1/(n+1)
Sn=1-1/2+1/2-1/3……-1/(n+1)
=1-1/(n+1)
(1)因为an+1=nan/(n+1)(nan+1)所以1/an+1=n+1+(n+1)/nan
所以d=bn+1-bn=1/(n+1)an+1-1/nan=[n+1+(n+1)/nan]/(n+1)-1/nan=1+1/nan-1/nan=1
而b1=1/a1=2,所以{bn}也是以b1=2为首项,为d=1为公差的等差数列
(2)请问SN是数列AN还是BN的和呢,在此算B...
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(1)因为an+1=nan/(n+1)(nan+1)所以1/an+1=n+1+(n+1)/nan
所以d=bn+1-bn=1/(n+1)an+1-1/nan=[n+1+(n+1)/nan]/(n+1)-1/nan=1+1/nan-1/nan=1
而b1=1/a1=2,所以{bn}也是以b1=2为首项,为d=1为公差的等差数列
(2)请问SN是数列AN还是BN的和呢,在此算BN的和吧,
由(1)知,bn是等差数列,sn=nb1+n(n-1)d/2=2n+n(n-1)/2=(n*n+3n)/2
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