1樓:kz菜鳥無敵
a4=a1加3d
s4=4a1加6d
a4加b4=-20
s4-b4=43
相加a4加s4=23
a1加3d加4a1加6d=23
9d=18
d=2a4=7,代入b4=-27,則q=-3an=2n-1,bn=-3^n
tn=-3^n(2n-1)
2樓:帥凝思
∵a4+b4=27,s4-b4=10 ∴a4+s4=37 ∴a4+2a1+2a4=37 ∴2a1+3a4=37
∴5a1+9d=37 ∴9d=27 ∴d=3 ∴an=a1+(n-1)d=3n-1
∵a4+b4=27 ∴11+2q³=27 ∴q³=8 ∴q=2 ∴bn=b1q^(n-1)=2^n
∵tn=anb1+an-1b2+...+a1bn ∴2tn=anb2+an-1b3+...+a2bn+a1bn+1
兩式相減得:tn=(an-an-1)b2+(an-1-an-2)b3+...+(a2-a1)bn+a1bn+1-anb1
=3(b2+b3+...+bn)+a1bn+1-anb1
=3×2²[2^(n-1)-1]+2×2^(n+1)-2an
=3×2×2^n-12+4×2^n-2an
=6bn-12+4bn-2an
∴tn=10bn-12-2an 即 tn+12=﹣2an+10bn
已知{an}是等差數列,其前n項的和為sn,{bn}是等比數列,且a1=b1=2,a4+b4=21,s4+b4=30.(1)求數列{an
3樓:城佳晨
(1)設等差數列的公差為d,等比數列的公比為q.由a1=b1=2,得a4=2+3d,b4=2q3,s4=8+6d.…(3分)
由條件a4+b4=21,s4+b4=30,得方程組2+3d+2q3=21
8+6d+2q3=30
解得d=1
q=2所以an=n+1,bn=2n,n∈n*.(2)由題意知,cn=(n+1)×2n.
記tn=c1+c2+c3+…+cn.
則tn=c1+c2+c3+…+cn
=2×2+3×22+4×23+…+n×2n-1+(n+1)×2n,2 tn=2×22+3×23+…+(n-1)×2n-1+n×2n+(n+1)2n+1,
所以-tn=2×2+(22+23+…+2n)-(n+1)×2n+1,即tn=n?2n+1,n∈n*.
已知{an}是等差數列,其前n項和為sn,{bn}是等比數列,且a1=b1=2,a4+b4=27,s4-b4=10.(ⅰ)求數列{an}
4樓:浮雲
(ⅰ)設等差數列的公差為d,等比數列的公比為q,
由a1=b1=2,得a4=2+3d,b4=2q3,s4=8+6d,
由條件a4+b4=27,s4-b4=10,
得方程組
2+3d+3q
=278+6d?2q
=10,
解得d=3
q=2,
故an=3n-1,bn=2n,n∈n*.
(ⅱ)方法一,由(ⅰ)得,tn=2an+22an-1+23an-2+…+2na1; ①;
2tn=22an+23an-1+…+2na2+2n+1a1; ②;
由②-①得,tn=-2(3n-1)+3×22+3×23+…+3×2n+2n+2
=12(1?n?1
)1?2
+2n+2-6n+2
=10×2n-6n-10.(n∈n*).
方法二:數學歸納法,
③當n=1時,t1+12=a1b1+12=16,-2a1+10b1=16,故等式成立,
④假設當n=k時等式成立,即tk+12=-2ak+10bk,
則當n=k+1時有,
tk+1=ak+1b1+akb2+ak-1b3+…+a1bk+1
=ak+1b1+q(akb1+ak-1b2+…+a1bk)
=ak+1b1+qtk
=ak+1b1+q(-2ak+10bk-12)
=2ak+1-4(ak+1-3)+10bk+1-24
=-2ak+1+10bk+1-12.
即tk+1+12=-2ak+1+10bk+1,因此n=k+1時等式成立.
③④對任意的n∈n*,tn+12=-2an+10bn成立.
∴tn=-2an+10bn-12=10×2n-6n-10.(n∈n*).
5樓:英韋本櫻花
(1)設數列的公差是d,的公比是q,依題意2+3d+2q^3=27,①
8+6d-2q^3=10,②
①+②,10+9d=37,d=3,
代入①,11+2q^3=27,q^3=8,q=2.
∴an=2+3(n-1)=3n-1,
bn=2^n.
(2)tn=2*2+5*2^2+8*2^3+……+(3n-1)*2^n,③
∴2tn=
2*2^2+5*2^3+……+(3n-4)*2^n+(3n-1)*2^(n+1),④
③-④,-tn=4+3(2^2+2^3+……+2^n)-(3n-1)*2^(n+1)
=4-3[2^2-2^(n+1)]-(3n-1)*2^(n+1),=-8-(3n-4)*2^(n+1),
∴tn=8+(3n-4)*2^(n+1),∴tn-8=ab.
已知數列{an}是等差數列,其前n項和為sn,{bn}是等比數列,且a1=b1=2,a4+b4=27,s4-b4=10.(1)求數列{
6樓:165內飾旱茨
(ⅰ)設等差數列的公差為d,等比數列的公比為q,由a1=b1=2,得a4=2+3d,b4=2q3,s4=8+6d,由a4+b4=27,s4-b4=10,得方程組2+3d+2q
=278+6d?2q
=10,
解得d=3
q=2,
所以:an=3n-1,bn=2n;
(ⅱ)由(ⅰ)知an?bn=(3n-1)?2n,則tn=2×2+5×22+8×23+…+(3n-1)×2n,①2tn=2×22+5×23+…+(3n-4)×2n+(3n-1)×2n+1,②
由①-②得,-tn=2×2+3(22+23+…+2n)-(3n-1)×2n+1
=4+3×4(1?n?1
)1?2
-(3n-1)×2n+1
=-(3n-4)×2n+1-8.
所以tn=(3n-4)×2n+1+8.
已知{an}是等差數列,其前n項和為sn,{bn}是等比數列,且a1=b1=2,a4+b4=27,s4-b4=10.(1)求數列{an}
7樓:10快樂
(1)設等差數列的公差為d,等比數列的公比為q,由a1=b1=2,得a4=2+3d,b4=2q3,s4=8+6d,由a4+b4=27,s4-b4=10,得方程組2+3d+2q
=278+6d?2q
=10,
解得d=3
q=2,
所以:an=3n-1,bn=2n.
(2)證明:由第一問得:tn=2×2+5×22+8×23+…+(3n-1)×2n; ①;
2tn=2×22+5×23+…+(3n-4)×2n+(3n-1)×2n+1,②.
由①-②得,-tn=2×2+3×22+3×23+…+3×2n-(3n-1)×2n+1
=6×(1?n
)1?2
-(3n-1)×2n+1-2
=-(3n-4)×2n+1-8.
即tn-8=(3n-4)×2n+1.
而當n≥2時,an-1bn+1=(3n-4)×2n+1.∴tn-8=an-1bn+1(n∈n*,n≥2).
已知{an}是等差數列,其前n項和為sn,{bn}是等比數列,且a1=b1=2,a4+b4=27,s4-b4=10
8樓:匿名使用者
tn=2an+22an-1+23an-2+…+2na1; ①;
2tn=22an+23an-1+…+2na2+2n+1a1; ②;
由②-①得,tn=-2(3n-1)+3×22+3×23+…+3×2n+2n+2
=12(1-2 n-1)
1-2+2n+2-6n+2
=10×2n-6n-10;
而-2an+10bn-12=-2(3n-1)+10×2n-12=10×2n-6n-10;
故tn+12=-2an+10bn(n∈n*).
9樓:不見幸村
a4+b4=27 (1) s4-b4=10 (2) (1)式和(2)式相加得 a4+s4=37
a1+a4=a2+a3 所以 s4=2(a1+a4) 所以 a4+2(a1+a4)=37 得3a4+2a1=37即3a4+4=37 得a4=11
由等差公式得an=3n-1
由上可知b4=16 所以公比為2 bn=2 q*(n-1)
10樓:匿名使用者
題目最好手寫拍下來,方便看得清
11樓:沈電火車俠
將兩式相加,得a4+s4=37,帶公式解得d=3,再代入第一個式子,解得q=2,所以an=3n-1,bn=2的n次冪,
已知{an}是等差數列,其前n項和為sn,{bn}是等比數列,且a1=b1=2,a4+b4=27,s4-b4=10
12樓:匿名使用者
(1)設數列的公差是d,的公比是q,依題意2+3d+2q^3=27,①
8+6d-2q^3=10,②
①+②,10+9d=37,d=3,
代入①,11+2q^3=27,q^3=8,q=2.
∴an=2+3(n-1)=3n-1,
bn=2^n.
(2)tn=2*2+5*2^2+8*2^3+……+(3n-1)*2^n,③
∴2tn= 2*2^2+5*2^3+……+(3n-4)*2^n+(3n-1)*2^(n+1),④
③-④,-tn=4+3(2^2+2^3+……+2^n)-(3n-1)*2^(n+1)
=4-3[2^2-2^(n+1)]-(3n-1)*2^(n+1),=-8-(3n-4)*2^(n+1),
∴tn=8+(3n-4)*2^(n+1),∴tn-8=ab.
13樓:真者降臨
∵a4+b4=27,s4-b4=10 ∴a4+s4=37 ∴a4+2a1+2a4=37 ∴2a1+3a4=37
∴5a1+9d=37 ∴9d=27 ∴d=3 ∴an=a1+(n-1)d=3n-1
∵a4+b4=27 ∴11+2q³=27 ∴q³=8 ∴q=2 ∴bn=b1q^(n-1)=2^n
∵tn=anb1+an-1b2+...+a1bn ∴2tn=anb2+an-1b3+...+a2bn+a1bn+1
兩式相減得:tn=(an-an-1)b2+(an-1-an-2)b3+...+(a2-a1)bn+a1bn+1-anb1
=3(b2+b3+...+bn)+a1bn+1-anb1
=3×2²[2^(n-1)-1]+2×2^(n+1)-2an
=3×2×2^n-12+4×2^n-2an
=6bn-12+4bn-2an
∴tn=10bn-12-2an 即 tn+12=﹣2an+10bn
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