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=](/uploads/image/z/18678424-40-4.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%3Dan%5E2%2Bbn%2Bc%2C%E8%AE%B0%E5%85%AC%E5%B7%AE%E4%B8%BAd%2C%E5%88%99a1%2Bd%2Bc%3D)
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.
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.
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