两个数列{an}和{bn}满足bn=a1+2a2+3a3+.+nan1+2+3+...+n (n€N+). ① 若{b}
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两个数列{an}和{bn}满足bn=a1+2a2+3a3+.+nan1+2+3+...+n (n€N+). ① 若{b}是等差数列,求证{a}也是等
两个数列{an}和{bn}满足bn=a1+2a2+3a3+.+nan/1+2+3+...+n (n€N+).
① 若{b}是等差数列,求证{a}也是等差数列
② 若{a}是等差数列,求证{b}也是等差数列
两个数列{an}和{bn}满足bn=a1+2a2+3a3+.+nan/1+2+3+...+n (n€N+).
① 若{b}是等差数列,求证{a}也是等差数列
② 若{a}是等差数列,求证{b}也是等差数列
![两个数列{an}和{bn}满足bn=a1+2a2+3a3+.+nan1+2+3+...+n (n€N+). ① 若{b}](/uploads/image/z/15754937-41-7.jpg?t=%E4%B8%A4%E4%B8%AA%E6%95%B0%E5%88%97%7Ban%7D%E5%92%8C%7Bbn%7D%E6%BB%A1%E8%B6%B3bn%3Da1%2B2a2%2B3a3%2B.%2Bnan1%2B2%2B3%2B...%2Bn+%28n%E2%82%ACN%2B%29.+%E2%91%A0+%E8%8B%A5%EF%BD%9Bb%EF%BD%9D)
1.写出bn和bn-1的表达式,把分母乘过去,两式相减,得到
(1+2+……+n-1)(bn-bn-1)=n(an-bn)
n(n-1)(bn-bn-1)/2=n(an-bn)
即an-bn=(n-1)(bn-bn-1)/2
带入bn=b1+(n-1)d
得an=b1+(n-1)(3d/2),等差
2.代入an=a0+nd
bn=[a0+d+2a0+2*2d+……+na0+n*nd]/[1+……+n]
=a0+d*[1^2+……+n^2]/[1+……+n]
=a0+d*[n(n+1)(2n+1)/6]/[n(n+1)/2]
=a0+d(2n+1)/3
显然也是等差数列
(1+2+……+n-1)(bn-bn-1)=n(an-bn)
n(n-1)(bn-bn-1)/2=n(an-bn)
即an-bn=(n-1)(bn-bn-1)/2
带入bn=b1+(n-1)d
得an=b1+(n-1)(3d/2),等差
2.代入an=a0+nd
bn=[a0+d+2a0+2*2d+……+na0+n*nd]/[1+……+n]
=a0+d*[1^2+……+n^2]/[1+……+n]
=a0+d*[n(n+1)(2n+1)/6]/[n(n+1)/2]
=a0+d(2n+1)/3
显然也是等差数列
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