已知等差数列{an}的前n项和Sn=an^2+bn+c,记公差为d,则a1+d+c=

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$已知等差数列{an}的前n项和Sn=an^2+bn+c,记公差为d,则a1+d+c=$

已知等差数列{an}的前n项和Sn=an^2+bn+c,记公差为d,则a1+d+c=
已知等差数列{an}的前n项和Sn=an^2+bn+c,记公差为d,则a1+d+c=

已知等差数列{an}的前n项和Sn=an^2+bn+c,记公差为d,则a1+d+c=
由{an}为等差数列,有Sn=a1n+[n(n-1)d/2]=a1n+(d/2)n^2-(d/2)n=(d/2)n^2+[a1-(d/2)]n,其中a1为首项,d为公差.
依题有Sn=an^2+bn+c,对照上式,有a=d/2,b=a1-(d/2),c=0,
故a1+d+c=a1+d=b+(d/2)+d=b+(3d/2)=b+3a.

a1=a+b+c
a1+a2=4a+2b+c
a2=3a+b
d=a2-a1=2a-c
a1+d+c=a+b+c+2a-c+c=3a+b+c

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