数列{an}满足an=n(n+1)^2,是否存在等差数列{bn}使an=1*b1+2*b2+3*b3+...n*bn,对于一切正整数恒成立,并证明
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数列{an}满足an=n(n+1)^2,是否存在等差数列{bn}使an=1*b1+2*b2+3*b3+...n*bn,对于一切正整数恒成立,并证明
设{Bn}公差为d
An=1*B1+2*B2+3*B3+...n*Bn
=B1+2(B1+d)+3(B1+2d)+4(B1+3d)+...+n[B1+(n-1)]d
=B1+2B1+3B1+...+nB1 + 2d+6d+12d+...+(n-1)nd
=B1*(1+2+3+...+n) + d*[1*2+2*3+...+(n-1)n]
=B1*n(n+1)/2 + d*(n-1)n(n+1)/3
因为 An=n(n+1)^2
所以 B1*n(n+1)/2 + d*(n-1)n(n+1)/3=n(n+1)^2
两边同时除以n(n+1),得:B1/2 + (n-1)d/3 = n+1
3B1 + 2(n-1)d = 6n+6
B1 + 2B1 + 2(n-1)d = 6n+6
B1 + 2Bn = 6n+6
所以 Bn=3n+1
即数列{Bn}是以4为首项,3为公差的等差数列
cn=a(n+1)-an=(n+1)*b(n+1)
c(n+1)-cn=(n+2)b(n+2)-(n+1)*b(n+1)=b(n+2)+(n+1)d=b(2n+3)=常数
d=0
bn公差为0
an=(1+2+3+...+n)*b1=n(n+1)^2*b1=n(n+1)^2
b1=1
所以bn=1
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