已知等差数列〔an〕的前4项和为10,且a2 ,a3 ,a7成等比数列?(1)试求数列〔an〕的通项公式an?(2)...已知等差数列〔an〕的前4项和为10,且a2 ,a3 ,a7成等比数列?(1)试求数列〔an〕的通项公式an?
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已知等差数列〔an〕的前4项和为10,且a2 ,a3 ,a7成等比数列?(1)试求数列〔an〕的通项公式an?(2)...已知等差数列〔an〕的前4项和为10,且a2 ,a3 ,a7成等比数列?(1)试求数列〔an〕的通项公式an?
已知等差数列〔an〕的前4项和为10,且a2 ,a3 ,a7成等比数列?(1)试求数列〔an〕的通项公式an?(2)...
已知等差数列〔an〕的前4项和为10,且a2 ,a3 ,a7成等比数列?(1)试求数列〔an〕的通项公式an?(2)设bn=2^an,求数列〔bn〕的前n项和Sn
已知等差数列〔an〕的前4项和为10,且a2 ,a3 ,a7成等比数列?(1)试求数列〔an〕的通项公式an?(2)...已知等差数列〔an〕的前4项和为10,且a2 ,a3 ,a7成等比数列?(1)试求数列〔an〕的通项公式an?
(a1+2d)²=(a1+d)×(a1+6d),得3a1d+2d²=0,则:d=0,此时a1=5/2,则an=5/2,Sn=(5/2)n;或者d=-(3/2)a1,则d=3,a1=-2,所以an=3n-5,Sn=n(3n-7)/2.
前4项和是a1+6d=10,a2 ,a3 ,a7成等比数列,(a1+2d)/(a1+d)=(a1+6d)/(a1+2d),解得a1=,d=
4a1+6d=10
(a1+2d)^2=(a1+d)(a1+6d) a1=-2 d=3
an=3n-5
bn=2^(3n-5) b1=1/4 q=8
Sn=1/4(1-8^n)/(1-8)
=1/28(8^n-1)
(1) a+(a+d)+(a+2d)+(a+3d) = 10
2a + 3d =5
(a+2d)^2 = (a+d)*(a+6d)
a =-2, d=3
an = -2+(n-1)*3 = 3n - 5
(2) bn = 2^(3n-5)
sn = 2^(-2) [ 1- 2^(3n)] / (1-2^3) = (2^(3n)-1)/28
(1).a2+a3=5, a2^2=(a2+d)*(a2+5d), 所以an=3n-5
(2)bn=2^an=2^(3n-5)=[8^(n-1)]/4, sn=(8^n-1)/28
1. 4a1+6d=10...(1)
(a1+2d)^2=(a1+d)*(a1+6d)...(2)
由(1)(2)式解得:a1=-2 ,d=3或a1=5/2 ,d=0
an=3n-5或an=5/2
2.bn=2^(3n-5) 则b1=1/4 , q=8
Sn=(1/28)*(8^n-1)
或bn=2^(5/2)
Sn=n*[2^(5/2)]
前4项和=2(2a3-d)=10,a3=(5+d)/2,(a3)²=a2a7,(5+d)²/4=(5-d)(5+9d)/4,
(5+d)²=(5-d)(5+9d),d=0,3,
d=0,an=5/2;d=3,an=3n-5
2.bn=2^(5/2),Sn=4n√2;bn=2^(3n-5),q=8,b1=1/4,Sn=(1/4)(1-8^n)/(1-8)=(8^n-1)/28