等比数列an中a1等于2
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C.充要,因为a1/a3=a5/a7=1/q^2,即从a1
a1=S1=2^1-1=1a2=S2-S1=2^2-1-1=2公比q=a2/a1=2/1=2an是等比数列——首项是1,公比是2an^2也是等比数列——首项是a1^2=1,公比是q^2=4a1^2+a
设{an}的公比为q,则a2=2q,a3=2q^2则(a2+1)^2=(a1+1)(a3+1)即(2q+1)^2=3(2q^2+1)解得q=1所以{an}为常数数列Sn=na1=2n
∵a1,1/2a3,2a2成等差数列∴2×1/2a3=a1+2a22即a3=a1+2a2∵{an}是等比数列,∴a1q²=a1+2a1q∴q²=1+2q,即q²-2q-1
解因为数列是等比数列,且公比为q则a2=a1qa3=a1q²又因为a1,1/2a3,2a2成等差数列所以有2*(1/2)a3=a1+2a2即a1q²=a1+2a1q即q²
(a2+1)²=(a1+1)(a3+1)a1=2,设an公比q(2q+1)²=3(2q²+1)4q²+4q+1=6q²+32q²-4q+2=
因数列{an}为等比,则an=2qn-1,因数列{an+1}也是等比数列,则(an+1+1)2=(an+1)(an+2+1)∴an+12+2an+1=anan+2+an+an+2∴an+an+2=2a
(Ⅰ)设{an}的公比为q,由已知得16=2q3,解得q=2.又a1=2,所以an=a1qn−1=2×2n−1=2n.(Ⅱ)由(I)得a2=8,a5=32,则b4=8,b16=32.设{bn}的公差为
这个图片不知道行不行啊再问:{an+1}为等比数列怎麽会有An+1+An-1=An再答:这是按照上面的公式算出来的啊,是等于2An因为an是等比数列,所以an+1*an-1=an*an
Sn=a1(1-q^n)/(1-q)S1=a1S2=a1(1+q)S3=a1(1+q+q^2)S2+2=a1(1+q)+2S3+2=a1(1+q+q^2)+2[a1(1+q+q^2)+2]*[a1+2
根据对数的运算性质,得log2a1+log2a2+…+log2a10=log2(a1a2a3…a9a10)=log2(a1a10)5=25,∴(a1a10)5=225,∵q=2,∴a1=14,∴a1+
高中数学老师的答案
填:216a₃²=a₂xa₄=a₁xa5=2x18=36∵a₁=2>0∴a₃=a₁q²>0∴a&
设等比数列{an}的公比为q,则可得an=2•qn-1,故an+1=2•qn-1+1,可得a1+1=3,a2+1=2q+1,a3+1=2q2+1,由于数列{an+1}也是等比数列,故(2q+1)2=3
an是等比设公比为qan^2也是等比公比是q^2Sn=2^n-1an=2^(n-1)公比是2a1=S1=1{an^2}是1为首项公比为4的等比数列和为(4^n-1)/3
S5=a1(1-q^5)/(1-q)=8(1-32)/(1-2)=248再问:前五项是什么再答:这个问题就是问你前五项的和S5,不需要回答前五项分别是什么
等比a3=a1q²a5=a3q²a7=a5q²所以(a3-a5+a7)/(a1-a3+a5)=q²=5/2同理aa9=a7q²所以(a5-a7+a9)
a1+a2+a3=a1(1+q+q^2)=2a4+a4+a6=a4(1+q+q^2)=4a4/a1=q^3=2a10+a11+a12=a1q^9(1+q+q^2)=(a1+a2+a3)(q^3)^2=
(a2+1)²=(a1+1)(a3+1)a1=2,设an公比q(2q+1)²=3(2q²+1)4q²+4q+1=6q²+32q²-4q+2=
设公比为q,a2²=a1*a3(a2+1)²=(a1+1)(a3+1)因为a1=2所以a2²=2a3(a2+1)²=3(a3+1)解得a2=2a3=2所以sn=