数列{ an}{ bn}满足关系式bn=1*a1+2*a2+3*a3…+nan/1+2+3+…+n,若{bn}为等差数列
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数列{ an}{ bn}满足关系式bn=1*a1+2*a2+3*a3…+nan/1+2+3+…+n,若{bn}为等差数列,求证数列{an}也是等差数
证:
设{bn}公差为d (d为常数).
a1/1=b1 a1=b1
bn=(a1+2a2+3a3+...+nan)/(1+2+3+...+n)
a1+2a2+3a3+...+nan=[n(n+1)/2]bn (1)
a1+2a2+3a3+...+(n+1)a(n+1)=[(n+1)(n+2)/2]b(n+1) (2)
(2)-(1)
(n+1)a(n+1)=[(n+1)(n+2)/2]b(n+1)-[n(n+1)/2]bn
2a(n+1)=(n+2)b(n+1)-nbn
=(n+2)(b1+nd)-n[b1+(n-1)d]
=nb1+n²d+2b1+2nd-nb1-n²d+nd
=3dn+2b1
a(n+1)=(3d/2)n+b1
a(n+1)-an=(3d/2)n+b1-(3d/2)(n-1)-b1=(3/2)d,为定值.
数列{an}是以b1为首项,(3/2)d为公差的等差数列.
设{bn}公差为d (d为常数).
a1/1=b1 a1=b1
bn=(a1+2a2+3a3+...+nan)/(1+2+3+...+n)
a1+2a2+3a3+...+nan=[n(n+1)/2]bn (1)
a1+2a2+3a3+...+(n+1)a(n+1)=[(n+1)(n+2)/2]b(n+1) (2)
(2)-(1)
(n+1)a(n+1)=[(n+1)(n+2)/2]b(n+1)-[n(n+1)/2]bn
2a(n+1)=(n+2)b(n+1)-nbn
=(n+2)(b1+nd)-n[b1+(n-1)d]
=nb1+n²d+2b1+2nd-nb1-n²d+nd
=3dn+2b1
a(n+1)=(3d/2)n+b1
a(n+1)-an=(3d/2)n+b1-(3d/2)(n-1)-b1=(3/2)d,为定值.
数列{an}是以b1为首项,(3/2)d为公差的等差数列.
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