设等比数列{an}的公比q=2,前n项和为Sn,s4除以a2-搜答案-神马作文网

S4=a1+a2+a3+a4=a2/q+a2+a2*q+a2*q^2S4/a2=1/q+1+q+q^2=7.5

设等比数列an的公比q=2,首项a1A4=a1*q^3=8a1S4=a1+a2+a3+a4=a1(1+q+q^2+q^3)=15a1S4/a4=15/8

qSn+a1=Sn+a(n+1)Sn=a1(1-q^n)/1-q=a1(2^n-1)S4/a2=a1*(2^4-1)/2a1=15/2

S4＝[a1(1－q^4)]/(1－q)＝[a1(1－2^4]/(1－2)＝15a1a2＝a1*q＝2a1∴S4/a2＝15a1/2a1＝15/2．

S4=a1(1-2^4)/(1-2)=15a1a2=a1·q=2a1S4/a2=15a1/(2a1)=15/2=7.5

S4=a1(1-2^n)/(1-2)=a1(2^n-1)a2=a1*2S4/a2=(2^n-1)/2再问：可以算出来吗再答：S4/a2=(2^n-1)/2=(2^4-1)/2=15/2

q=2等比数列a2=2a1s4=[a1/(1-2)]*(1-2&4)=(-a1)*(-15)=15a1s4/a2=15a1/2a1=15/2

S4=a1+a2+a3+a4S2=a1+a2所以S4/S2=1+q^2=1+4=5

第二题：1/(X-1)=1X>=2所以不等式解集为X=2第一题公比q若为正数的话,哪么应该大于1,因为要是q

S4=a1(1-q^4)/(1-q)=5a1(1-q^2)/(1-q)1+q^2=5q^2=4因为q

s4/a4=[a1(1-q^4)/(1-q)]/a1q^3=[(1-q^4)/(1-q)]/q^3=[(1-q)(1+q)(1+q^2)]/(1-q)]/q^3=(1+q)(1+q^2)/q^3=(1

s4=a1(1-q^4)/(1-q)a4=a1q^3s4/a4=(1-q^4)/q^3(1-q)=(1-1/16)/(1/16)=15再问：

Sn=a1(1-q^n)/1-qS4=a1*/(1-1/2)a4=a1*q^3S4/a4=/(1/2)*1/(1/2)^3=(1-1/16)/1/16=(15/16)*16=15*是乘号的意思a^b就

用等比数列的通项公式和求和公式S4=a1(1-q^4)/1-qa4=a1.q^3把q等于1/2带进去,就可以求出答案是15

由等比数列的求和公式和通项公式可得：S4a3=a1(1-24)1-2a1•22=15a14a1=154故答案为：154

∵{an+c}是等比数列∴（a1+c）（a3+c）=（a2+c）2即a1a3+c（a1+a3）+c2=a22+2a2c+c2∵a1a3=a22∴（a1+a3）c=2a2c即a1c（1+q2）=2a1q

首先得求的a1a4=5s2...a1q^3=5(a1+a1q)又.a3=a1q^2=2...所以.2q=5(a1+a1q)得.a1=(2q)/(5(1+q))又因为.a3=a1q^2=2得.q=1.2

等比数列an=a1*q^(n-1),Sn=a1(1-q^n)/(1-q)∴a3=2=a1*q^(3-1)=a1*q^2S4=5S2=>a1(1-q^4)/(1-q)=5*a1(1-q^2)/(1-q)

S4=a1(1-q4)/(1-q),S2=a1(1-q2)/(1-q),已知S4=5S2,则a1(1-q4)/(1-q)=5a1(1-q2)/(1-q),即q=±2,又公比q

S4=a1(1-q4)/(1-q),S2=a1(1-q2)/(1-q),已知S4=5S2,则a1(1-q4)/(1-q)=5a1(1-q2)/(1-q),即q=±2,又公比q