问题
解答题
已知数列{an}是等差数列,{bn}是等比数列,且a1=b1=2,b4=54,a1+a2+a3=b2+b3.
(1)求数列{an}和{bn}的通项公式;
(2)数列{cn}满足cn=anbn,求数列{cn}的前n项和Sn.
答案
(1)设{an}的公差为d,{bn}的公比为q
由b4=b1q3=54,得q3=
=27,从而q=354 2
因此bn=b1 • qn-1=2 • 3n-1(3分)
又a1+a2+a3=3a2=b2+b3=6+18=24,∴a2=8
从而d=a2-a1=6,故an=a1+(n-1)•6=6n-4(6分)
(2)cn=anbn=4 • (3n-2) • 3n-1
令Tn=1×30+4×31+7×32+…+(3n-5) • 3n-2+(3n-2) • 3n-1
3Tn=1×31+4×32+7×33+…+(3n-5) • 3n-1+(3n-2) • 3n(9分)
两式相减得-2Tn=1+3×31+3×32+3×33+…+3×3n-1-(3n-2) • 3n
=1+3 •
-(3n-2)•3n=1+3 • (3n-1-1) 3-1
-(3n-2) • 3n9(3n-1-1) 2
∴Tn=
+7 4
,又Sn=4Tn=7+(6n-7) • 3n(12分).3n(6n-7) 4