Question

等比数列（an）的各项均为正数，且a5*a6+a4*a7=18，则log3a1+log3a2+。。。。。+log3a9+log3a10=？

Answer 1

解:
由于{an}为等比数列
则:a5a6=a4a7=a3a8=a2a9=a1a10

又a5a6+a4a7=18
则:
2a5a6=18
a5a6=9

则:
log3(a1)+log3(a2)+...+log3(a9)+log3(a10)
=log3[a1*a2*a3*...*a10]
=log3[(a1a10)*(a2a9)*...*(a5a6)]
=log3[9*9*...*9]
=log3[9^5]
=log3[3^10]
=10log3[3]
=10

Answer 2

a5*a6+a5/q*a6*q=2a5*a6=18,所以a5*a6=9
log3a1+log3a2+。。。。。+log3a9+log3a10=log3a1a2...a9
=log3a1*a1q*a1q＾2*a1q＾8
=log3a1＾10*q＾45
=5log3a1＾2*q＾9
=5log3a5a6=5log3 9=10

Answer 3

aq^4*aq^5+aq^3*aq^6=18
2a^2q^9=18
a^2q^9=9

3底log a1+3底log a2+...+3底log a10＝
=3底log(a1*a2*……*a10)
=3底loga^10*q^(1+2+……+9)
=3底loga^10q^45
=3底log(a2q^9)^5
=5*3底log9
=5*2
=10