数列{ an}{ bn}满足关系式bn=1a1+2a2+3*a3…+nan/1+2+3+…+n

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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{ an}为等差数列，求证数列{ bn}也是等差数列

证明：若{ an}为等差数列,设公差为d,则
b[n]={(1+2+……+n)*a1+[1*2+2*3+……+(n-1)n]d}/(1+2+……+n)
=a1+(1^2+2^2+3^2+……+n^2)d-(1+2+……+n)d/(1+2+……+n)
=a1+[n(n+1)(2n+1)d/6-n(n+1)d/2]/[n(n+1)/2]
=a1+(2n+1)d/3-d
=a1+(2n-2)d
a1+(n-1)*2d
显然,b[n]是以a1为首项,2d为公差的等差数列.命题得证.

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