Question

1 在等比数列｛an}中,已知Sn=48,S2n=60,求S3n

Answer 1

(1)

Sn=48=a1+a2+……+an=48

S2n=a1+a2+……an+a(n+1)+……a(2n)=60

Sn+Sn*q^n=S2n

q^n=1/4

S3n=a1+……a(3n)=S2n+a(2n+1)+……a(3n)=S2n+Sn*q^2n=60+48*(1/4)^2=63

(2)

a(n+1)=2(an)+3

a(n+1)+3=2(an)+6=2(an+3)

a1+3=4

所以{an+3}是以4为首项,2为公比的等比数列

an+3=2^(n+1)

an=2^(n+1)-3