(1 2)设an是等差数列,bn=1 2的an次方,已知-
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/27 12:46:02
解题思路:考查了等差数列、等比数列的通项公式,以及二次函数的最值解题过程:
(a1)(b1)=1,因b1=1,则:a1=1则:(a2)(b2)=(a1+d)[b1q]=(1+d)q=4,则:(1+d)²q²=16(a3)(b3)=(a1+2d)[b1q
∵{An}是等差数列∴An-A(n-1)=d(d为公差)∵Bn=kAn+m∴B(n-1)=kA(n-1)+m∴Bn-B(n-1)=kAn+m-[kA(n-1)+m]=k[An-A(n-1)]=kd这个
设等差数列的等差为d,等比数列的等比是q,由a3=b3,得a4-d=b4q,又∵a4=b4,∴a4-a4q=d,∵S5-S3T4-T2=7,∴a5+a4b4+b3=a4+d+a4a4+a4q=7,即3
证:由数列{an}是等差数列,得an=a1+(n-1)d,其中a1为首项,d为公差.b1b2b3=[(1/2)^(a1)][(1/2)^(a1+d)][(1/2)^(a1+2d)]=(1/2)(a1+
设等差数列的等差为d,等比数列的等比是q则a3=b3a4-d=b4/q又∵a4=b4∴a4-d=a4/qa4-a4/q=d∵(S5-S3)/(T4-T2)=5∴(a5+a4)/(b4+b3)=(a4+
(1/2)^a1+(1/2)^a2+(1/2)a^3=21/8(1/2)^a1*(1/2)^a2*(1/2)^a3=1/8(1/2)^(a1+a2+a3)=1/8a2=1a1=1或a1=4a3=4或1
a(n+1)=√[bn*b(n+1)]2bn=an+an+12bn=√[bn*b(n-1)]+√[bn*b(n+1)]2√bn=√b(n-1)+√b(n+1)所以数列{√bn}为等差数列√b1=√2(
a1+a2+a3+…+an=na1+[n(n-1)d]/2,则bn=a1+(d/2)(n-1),从而b(n+1)-bn=[a1+(d/2)n]-[a1+(d/2)(n-1)]=d/2=常数,则数列{b
LZbn的通项公式求错了,bn=4n-2而不是bn=4n-1;你验证下b1就知道了所以1/anbn=1/[2*(2n-1)(2n+1)]=1/4*[1/(2n-1)-1/(2n+1)]所以1/a1b1
首先等差数列的通项公式是关于n的一次式bn是等差数列,设bn=A*n+B则:a1+a2+a3+a4+...+an=n(A*n+B)=A(n^2)+Bna1+a2+a3+a4+...+a(n-1)=A(
(1)证明:an-2=2-4/a(n-1)=(2a(n-1)-4)/a(n-1)1/(an-2)=a(n-1)/(2a(n-1)-4)=1/2*a(n-1)/(a(n-1)-2)=1/2[1+2/(a
a1,a3,a4成等比数列a3=a1+(3-1)*(-2)=a1-4;a4=a1-6a3*a3=a1*a4(a1-4)(a1-4)=a1(a1-6)a1*a1-8a1+16=a1*a1-6a12a1=
设Sn=k(7n^2+n)an=Sn-S(n-1)=k(14n-6)Tn=k(4n^2+27n)bn=Tn-T(n-1)=k(8n+23)an:bn==(14n-6)/(8n+23)再问:错·再答:哪
设an=a1+(n-1)d,bn=an+a(n-1)=a1+(n-1)d+a1+nd=2a1+(2n-1)dbn为首项为2a1-d,公差为2d的等差数列
B978an=2^(n-1)bn=1-nC10=A10+B10=978
(1)因为{an+1-an}是等差数列,所以a2-a1=-2,a3-a2=-1,a4-a3=0,…,an-an-1=n-4,以上各式相加得,an-a1=(n−1)(n−6)2,即an=6+(n−1)(
设等差数列{an}的公差为d,则an=a1+(n-1)d.∴bn=(12)a1+(n-1)db1b3=(12)a1•(12)a1+2d=(12)2(a1+d)=b22.由b1b2b3=18,得b23=
设bn的公比为q,首项为bb+bq+bq^2=21/8b^3q^3=1/8所以bq=1/2解得b=1/8,q=4b=2,q=1/4当b=1/8,q=4,则d=-2,a1=3,an=5-2n当b=2,q
很高兴为您答题,如果有其他需要帮助的题目,您可以求助我.