作业帮数列 (30 20:12:4)设两个数列{An},{Bn}满足Bn=（a1+2*a2+3*a3+…+n*an）/（1+2+3+…+n）,若{Bn}为等差数列,求证：{An}也为等差数列
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数列 (30 20:12:4)设两个数列{An},{Bn}满足Bn=（a1+2*a2+3*a3+…+n*an）/（1+2+3+…+n）,若{Bn}为等差数列,求证：{An}也为等差数列
数列 (30 20:12:4)
设两个数列{An},{Bn}满足Bn=（a1+2*a2+3*a3+…+n*an）/（1+2+3+…+n）,若{Bn}为等差数列,求证：{An}也为等差数列
数列 (30 20:12:4)设两个数列{An},{Bn}满足Bn=（a1+2*a2+3*a3+…+n*an）/（1+2+3+…+n）,若{Bn}为等差数列,求证：{An}也为等差数列
证明:
若数列{Bn}是等差数列,
则:设公差为d,
则有Bn=B1+(n-1)d,
由:
Bn=（A1+2*A2+3*A3+…+n*An)/(1+2+3+…+n),
可知:
A1+2*A2+3*A3+…+n*An
=(1+2+3+…+n)Bn
=n(n+1)/2*Bn,
所以
A1+2*A2+3*A3+…+(n-1)*A(n-1)
=(n-1)n/2*B(n-1),
两式相减,得
n*An=n(n+1)/2*Bn-(n-1)n/2*B(n-1)
所以:
An
=(n+1)/2*Bn-(n-1)/2*B(n-1)
=Bn+(n-1)/2*(Bn-B(n-1))
=B1+(n-1)d+(n-1)/2*d
=B1+(n-1)*(3d/2),
故{An}也是等差数列.