已知a 00101010b和b 40d
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=$B4*C4+$B5*C5
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+
(1)设{an}的公差为d,{bn|的公比为q∵a1=b1=2,∴a4+b4=2+3d+2q^3=27s4-b4=8+6d-2q^3=10相加10+9d=37∴d=3,q^3=8,q=2∴an=2+3
解1.a(4)+b(4)=27s(4)-b(4)=10,a(4)+s(4)=37=5a(1)+9da(4)-s(4)+2b(4)=17=a(1)+3d-4a(1)-6d+2b(4)=-3a(1)-3d
1.1)设a1=b1=t,依题意t+3d=t*d^3,(1)t+9d=t*d^9,(2)由(1)3d/t=d^3-1代入(2)有d^9-3d^3+2=0(d^3+2)*(d^3-1)^2=0于是d^3
S9/T9=9a5/9b5=a5/b5=63/12=21/4S8/T8=4(a4+a5)/[4(b4+b5)]=(a4+a5)/(b4+b5)=56/11S7/T7=7a4/7b4=a4/b4=49/
an=a1+(n-1)dbn=b1*d^(n-1)因a1=b1,a4=b4a1+3d=a1*d^3,a1=3d/(d^3-1)又a10=b10,a1=9d/(d^9-1)∴1/(d^3-1)=3/(d
(1)当n=1时,a1=S1=2a1-1,∴a1=1…(1分)当n≥2时,an=Sn-Sn-1=(2an-1)-(2an-1-1)=2an-2an-1,即 anan−1=2…(3分)∴数列{
∵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∴
设第一个的公差是d1,第二个公差是d2则a5-a1=4d1b5-b2=3d2a1=b2,a5=b5所以4d1=3d2d2/d1=4/3所以b6-b1=5d2a3-a2=d1所以原式=5d2/d1=5×
(1)设公差为d,公比为q,由题意得1+3d+q3=−204+6d−q3=43,解之得:d=2q=−3,从而an=2n−1,bn=(−3)n−1.…(5分)(2)Tn=1•(−3)0+3•(−3)1+
(Ⅰ)设等差数列的公差为d,等比数列的公比为q,由a1=b1=2,得a4=2+3d,b4=2q3,s4=8+6d,由条件a4+b4=27,s4-b4=10,得方程组2+3d+3q3=278+6d−2q
根据题意得,5+a=3,4=b+1,解得a=-2,b=3,所以,3b4-6a3b-4b4+2ba3=3×34-6×(-2)3×3-4×34+2×3×(-2)3=243+144-324-48=387-3
4=b1*q^3=2*q^3=54q^3=27q=3bn=2*3^(n-1)b2=b1*q=2*3=6b3=b1*q^2=2*9=18a1+a2+a3=b2+b3a1+a1+d+a1+2d=6+183
1080.由题目可知:B中任何一个元素在A中都有一个或两个元素与它对应,B中有4个元素,那么解题思路就是将A中的六个元素分为4组,每组1至2个元素,然后将每一组分别与B中的一个元素相对应.因此只要求出
a+b+c=0,两边平方得:a2+b2+c2+2ab+2bc+2ca=0,∵a2+b2+c2=1,∴1+2ab+2bc+2ca=0,∴ab+bc+ca=-12;ab+bc+ca=-12两边平方得:a2
(1)由已知Sn=n2,得a1=S1=1当n≥2时,an=Sn-Sn-1=n2-(n-1)2=2n-1所以an=2n-1(n∈N*)由已知,b1=a1=1设等比数列{bn}的公比为q,由2b3=b4得
虽然不知道你b4-s3=?,但是两个等差数列a1、b1已知,只有d1,d2两个未知数,由a2+b4=21,b4-s3=,即可得d1+3*d2=17,3*d2-3*d1=?解出即可
a4=a1+3db4=b1*d^3所以a1+3d=a1*d^3a1=3d/(d^3-1)a10=a1+9db10=b1*d^9所以a1+9d=a1*d^9a1=9d/(d^9-1)所以3d/(d^3-
1.a1=b1a4=b4a1+3d=b1*d^3=a1*d^33d=a1(d^3-1)(1)a10=b10a1+9d=b1*d^9=a1*d^99d=a1(d^9-1)=a1(d^3-1)(d^6+d