已知数列{an}满足an-1-2an+an+1=0(n∈N*且n≥2),且a1=2,a3=4.数列{bn}的前n项和为S
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已知数列{an}满足an-1-2an+an+1=0(n∈N*且n≥2),且a1=2,a3=4.数列{bn}的前n项和为Sn=2bn-1(n∈N*).
(1)求数列{an},{bn}的通项公式;
(2)符号[x]表示不超过实数x的最大整数,记cn=[log2(an-1)],Tn为数列{cn}的前n项和,求T
(1)求数列{an},{bn}的通项公式;
(2)符号[x]表示不超过实数x的最大整数,记cn=[log2(an-1)],Tn为数列{cn}的前n项和,求T
(1)∵数列{an}满足an-1-2an+an+1=0(n∈N*且n≥2),
∴数列{an}是等差数列,设公差为d,
∵a1=2,a3=4.∴a3-a1=2d=4-2,解得d=1.
∴an=2+(n-1)=n+1.
由数列{bn}的前n项和为Sn=2bn-1(n∈N*).
当n≥2时,bn=Sn-Sn-1=(2bn-1)-(2bn-1-1),化为bn=2bn-1.
当n=1时,b1=2b1-1,b1=1.
∴数列{bn}是等比数列,∴bn=2n-1.
(2)由(1)知an=n+1,∴cn=[log2n].
当2k≤n<2k+1时,[log2n]=k,k∈N.
∴T2n=[log21]+[log22]+…+[log2(2n-1)]+[log22n]
=([log221]+[log23])+([log222]+…+[log27])+([log223]+…+[log215])+…+([log22n-1]+[log2(2n-1+1)]+…+[log2(2n-1)])+[log22n],
∴T2n=2+2×22+3×23+…+(n-1)2n-1+n,
2T2n=1×22+2×23+…+(n-2)2n-1+(n-1)2n+2n,
两式相减得:-T2n=2+22+…+2n-1-n-(n-1)2n,
∴T2n=(n-2)2n+n+2.
∴数列{an}是等差数列,设公差为d,
∵a1=2,a3=4.∴a3-a1=2d=4-2,解得d=1.
∴an=2+(n-1)=n+1.
由数列{bn}的前n项和为Sn=2bn-1(n∈N*).
当n≥2时,bn=Sn-Sn-1=(2bn-1)-(2bn-1-1),化为bn=2bn-1.
当n=1时,b1=2b1-1,b1=1.
∴数列{bn}是等比数列,∴bn=2n-1.
(2)由(1)知an=n+1,∴cn=[log2n].
当2k≤n<2k+1时,[log2n]=k,k∈N.
∴T2n=[log21]+[log22]+…+[log2(2n-1)]+[log22n]
=([log221]+[log23])+([log222]+…+[log27])+([log223]+…+[log215])+…+([log22n-1]+[log2(2n-1+1)]+…+[log2(2n-1)])+[log22n],
∴T2n=2+2×22+3×23+…+(n-1)2n-1+n,
2T2n=1×22+2×23+…+(n-2)2n-1+(n-1)2n+2n,
两式相减得:-T2n=2+22+…+2n-1-n-(n-1)2n,
∴T2n=(n-2)2n+n+2.
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