设Sn为数列an的前n项和,Sn=kn∧2+n+r,n∈N*,(k是常数).(1)若an为等差数列,求r的值.(2)若r
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设Sn为数列an的前n项和,Sn=kn∧2+n+r,n∈N*,(k是常数).(1)若an为等差数列,求r的值.(2)若r=0且a2m(下标,后同)、a4m、a8m(m∈N*)成等比数列,求k的值.
a(1)=s(1)=k+1+r,
a(n+1)=s(n+1)-s(n)=k(2n+1)+1=2k(n+1-1) + k+1,
(1),
a(n+1)=2k(n+1-1)+k+1,a(1)=k+1+r.
a(2)=2k(1+1-1)+k+1=2k+k+1,a(1)=k+1+r,
2k=a(2)-a(1)=2k-r,r=0.
(2),
a(1)=k+1,
a(n+1)=2kn+k+1,
a(n)=2k(n-1)+k+1.
[a(4m)]^2=[2k(4m-1)+k+1]^2=a(2m)a(8m)=[2k(2m-1)+k+1][2k(8m-1)+k+1]=[2k(4m-1)+k+1-2k(2m)][2k(4m-1)+k+1+2k(4m)]=[2k(4m-1)+k+1]^2+2k(2m)[2k(4m-1)+k+1]-2k(2m)*(2k)(4m),
0=2k(2m)[2k(4m-1)+k+1] - 32(km)^2
=4km[2k(4m-1)+k+1 - 8km]
=4km[8km-2k+k+1-8km]
=4km[1-k]
0=k[1-k],
k=0或,k=1.
a(n+1)=s(n+1)-s(n)=k(2n+1)+1=2k(n+1-1) + k+1,
(1),
a(n+1)=2k(n+1-1)+k+1,a(1)=k+1+r.
a(2)=2k(1+1-1)+k+1=2k+k+1,a(1)=k+1+r,
2k=a(2)-a(1)=2k-r,r=0.
(2),
a(1)=k+1,
a(n+1)=2kn+k+1,
a(n)=2k(n-1)+k+1.
[a(4m)]^2=[2k(4m-1)+k+1]^2=a(2m)a(8m)=[2k(2m-1)+k+1][2k(8m-1)+k+1]=[2k(4m-1)+k+1-2k(2m)][2k(4m-1)+k+1+2k(4m)]=[2k(4m-1)+k+1]^2+2k(2m)[2k(4m-1)+k+1]-2k(2m)*(2k)(4m),
0=2k(2m)[2k(4m-1)+k+1] - 32(km)^2
=4km[2k(4m-1)+k+1 - 8km]
=4km[8km-2k+k+1-8km]
=4km[1-k]
0=k[1-k],
k=0或,k=1.
设Sn为数列an的前n项和,Sn=kn∧2+n+r,n∈N*,(k是常数).(1)若an为等差数列,求r的值.(2)若r
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