设Sn为数列{an}的前n项和,Sn=kn^2+n,n属于N*,其中k是常数若{an}为等差数列求r值2.若r=0且a2m,a4m,a8m(m属于N*)成等比数列,求k值
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![设Sn为数列{an}的前n项和,Sn=kn^2+n,n属于N*,其中k是常数若{an}为等差数列求r值2.若r=0且a2m,a4m,a8m(m属于N*)成等比数列,求k值](/uploads/image/z/6915414-30-4.jpg?t=%E8%AE%BESn%E4%B8%BA%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2CSn%3Dkn%5E2%2Bn%2Cn%E5%B1%9E%E4%BA%8EN%2A%2C%E5%85%B6%E4%B8%ADk%E6%98%AF%E5%B8%B8%E6%95%B0%E8%8B%A5%7Ban%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E6%B1%82r%E5%80%BC2.%E8%8B%A5r%3D0%E4%B8%94a2m%2Ca4m%2Ca8m%EF%BC%88m%E5%B1%9E%E4%BA%8EN%2A%EF%BC%89%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E6%B1%82k%E5%80%BC)
设Sn为数列{an}的前n项和,Sn=kn^2+n,n属于N*,其中k是常数若{an}为等差数列求r值2.若r=0且a2m,a4m,a8m(m属于N*)成等比数列,求k值
设Sn为数列{an}的前n项和,Sn=kn^2+n,n属于N*,其中k是常数若{an}为等差数列求r值
2.若r=0且a2m,a4m,a8m(m属于N*)成等比数列,求k值
设Sn为数列{an}的前n项和,Sn=kn^2+n,n属于N*,其中k是常数若{an}为等差数列求r值2.若r=0且a2m,a4m,a8m(m属于N*)成等比数列,求k值
等差数列求和通式为:Sn=n[a1+a1+(n-1)]/2=n(a1-1/2)+n^2/2
与Sn=kn^2+n比较,可知:k=1/2,a1-1/2=k =>a1=1
设公差为d,an=1+(n-1)d
a2m/a4m=a4m/a8m => a4 * a4 = a2 * a8a2=1+da4=1+3d
a8=1+7d
所以:(1+3d)(1+3d)=(1+d)(1+7d)
1+6d+9d^2=1+8d+7d^2
2d^2=2d => d=1 或 d=0,所以an是自然数数列或全1数列.
等差数列求和通式为: Sn=n[a1+a1+(n-1)]/2=n(a1-1/2)+n^2/2
与Sn=kn^2+n比较,可知:k=1/2, a1-1/2=k =>a1=1
设公差为d, an=1+(n-1)d
a2m/a4m=a4m/a8m => a4 * a4 = a2 * a8
a2=1+d
a4=1+3d
a8=1+7d
所...
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等差数列求和通式为: Sn=n[a1+a1+(n-1)]/2=n(a1-1/2)+n^2/2
与Sn=kn^2+n比较,可知:k=1/2, a1-1/2=k =>a1=1
设公差为d, an=1+(n-1)d
a2m/a4m=a4m/a8m => a4 * a4 = a2 * a8
a2=1+d
a4=1+3d
a8=1+7d
所以: (1+3d)(1+3d)=(1+d)(1+7d)
1+6d+9d^2=1+8d+7d^2
2d^2=2d => d=1 或 d=0, 即an是自然数数列或全1数列。
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题中没看到有r这个变量
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