已知{an}为等比数列,a1=1,a5=256,S n为等差数列{bn}的前n项和,b1=2,5S5=2S8 设Tn=a1b1+a2b2+...+anbn,求Tn
\u5df2\u77e5\u6570\u5217{an}\u662f\u7b49\u5dee\u6570\u5217\uff0c\u6570\u5217{bn}\u662f\u7b49\u6bd4\u6570\u5217\uff0c\u4e14\u5bf9\u4efb\u610f\u7684n\u2208N*\uff0c\u90fd\u6709a1b1+a2b2+a3b3+\u2026+anbn=n?2n+3\uff0e\uff08\uff08\u2160\uff09\u56e0\u4e3aa1b1+a2b2+a3b3+\u2026+anbn=n?2n+3\uff0c\u6240\u4ee5\u5f53n\u22652\u65f6\uff0ca1b1+a2b2+a3b3+\u2026+an-1bn-1=\uff08n-1\uff09?2n+2\uff0c\u4e24\u5f0f\u76f8\u51cf\uff0c\u5f97anbn=n?2n+3-\uff08n-1\uff09?2n+2=\uff08n+1\uff09?2n+2\uff0c\u800c\u5f53n=1\u65f6\uff0ca1b1=16\uff0c\u9002\u5408\u4e0a\u5f0f\uff0c\u4ece\u800canbn=\uff08n+1\uff09?2n+2\uff0c\u2026\uff083\u5206\uff09\u53c8\u56e0\u4e3a{bn}\u662f\u9996\u9879\u4e3a4\uff0c\u516c\u6bd4\u4e3a2\u7684\u7b49\u6bd4\u6570\u5217\uff0c\u5373bn=2n+1\uff0c\u6240\u4ee5an=2n+2\uff0c\u2026\uff084\u5206\uff09\u4ece\u800c\u6570\u5217{an+bn}\u7684\u524dn\u9879\u548cSn=n(4+2n+2)2+4(1?2n)1?2=2n+2+n2+3n-4\uff1b\u2026\uff086\u5206\uff09\uff08\u2161\uff09\u56e0\u4e3aan=4n+4\uff0canbn=\uff08n+1\uff09?2n+2\uff0c\u6240\u4ee5bn\uff1d2n\uff0c\u2026\uff0e\uff088\u5206\uff09\u5047\u8bbe\u6570\u5217{bn}\u4e2d\u7b2ck\u9879\u53ef\u4ee5\u8868\u793a\u4e3a\u8be5\u6570\u5217\u4e2d\u5176\u5b83r\uff0cr\u2208N\uff0cr\u22652\uff09\u9879bt1\uff0c\u2026\uff0cbtr\uff0c\uff08t1\uff1ct2\uff1c\u2026\uff1ctr\uff09\u7684\u548c\uff0c\u5373bk=bt1+\u2026+btr\uff0c\u4ece\u800c2k=2t1+\u2026+2tr\uff0c\u6613\u77e5k\u2265tr+1\uff0c\uff08*\uff09 \u2026\uff089\u5206\uff09\u53c82k=2t1+\u2026+2tr\u22642+22+\u2026+2tr=2(1?2tr)1?2=2tr+1-1\uff1c2tr+1\uff0c\u6240\u4ee5k\uff1ctr+1\uff0c\u6b64\u4e0e\uff08*\uff09\u77db\u76fe\uff0c\u4ece\u800c\u8fd9\u6837\u7684\u9879\u4e0d\u5b58\u5728\uff0e \u2026\uff0812\u5206\uff09
\u89e3\uff1a1) {an}\u4e3a\u7b49\u6bd4\u6570\u5217,a1=1,a5=256,\u4ee4an=a1q^(n-1)=q^(n-1),
=> q=(a5/a1)^(1/4)=\u00b14
q=4\u65f6\uff0can=4^(n-1);
q=-4\u65f6\uff0can=(-4)^(n-1).
2) \u4ee4bn=b1+(n-1)d=2+(n-1)d,
=> Sn=b1n+n(n-1)d/2=2n+n(n-1)d/2,
=> S5=10+10d, S8=16+28d,
5S5=2S8,
=> d=3,
=> bn=2+3(n-1)=3n-1
5S5=2S8 推出：5（b1+b5）*5=2（b1+b8）*8
25b1+25b5=16b1+16b8
9b1=16b8-25b5
得公差12d=18b1 d =3 bn=3n-1 n∈N*
an*bn=（3n-1）*4^(n-1) n∈N* 又称等差比数列，求和解法是，Tn 乘以个公比 做减法
Tn=2*1+5*4++8*4²……+（3n-1）*4^(n-1) --------------------①
4Tn=2*4+5*4²+……+（3n-1）*4^n ————————--------②
② - ① （注意 次幂相同的 想减）
3Tn= -2 -3[4+4²+4^3+4^(n-1)]+(3n-1）*4^n
整理，得Tn=n*4^n - (2/3)*4^n +(2/3) n∈N*
设公比为q,公差为d,
a5=a1q^4=q^4=256,q=土4.
由5S5=2S8 得
5[10+10d]=2[16+28d],
50+50d=32+56d,18=6d,d=3.
bn=3n-1.
Tn=2a1+5a2+……+（3n-1)an,
qTn=.....2a2+……+（3n-4)an+(3n-1)a<n+1>,
相减得（1-q)Tn=-a1+3(a1+a2+……+an)-(3n-1)a<n+1>
=3(1-q^n)/(1-q)-1-(3n-1)q^n
=[3-3*q^n-1-(3n-1)q^n+q+(3n-1)q^(n+1)]/(1-q)
=[2+q-(3n+2)q^n+(3n-1)q^(n+1)]/(1-q),
∴Tn=[2+q-(3n+2)q^n+(3n-1)q^(n+1)]/(1-q)^2,
q=4时Tn=[6-(3n+2)*4^n+(3n-1)*4^(n+1)]/9
=[2-(3n-2)*4^n]/3;
q=-4时Tn=[-2-(3n+2)*(-4)^n+(3n-1)*(-4)^(n+1)]/25
=[-2-(15n-2)*(-4)^n]/25.
∵{an}为等比数列,a1=1,a5=256,
∴a1*q^4=256
∴q=4
∴an=4^n-1
又∵{bn}为等差数列，b1=2,5S5=2S8
Sn=na1+n(n-1)/2*d
∴d=3
∴bn=3n-1
∴Tn=2*4^0+5*4^1+8*4^2+...+(3n-1)*q^n-1①
将等式两边同乘以4
4Tn=2*4^1+5*4^2+...+(3n-1)*q^n②
由①-②，得到-3Tn=2*4^0+3*4^1+3*4^2+...+(3n-3)*4^n-1-(3n-1)*4^n
∴-3Tn=2*4^0+3(4^1+4^2+...+4^n)-(3n-1)*4^n
∴-3Tn=2*4^0+4^(n-1)-4-(3n-1)*4^n
∴Tn=-2/3+4^(n-1)/3-(3n-1)*4^n/3
O(∩_∩)O，希望对你有帮助
{an}为等比数列,a1=1, a5=256
==>q^4=256,q=4
an=4^(n-1)
等差数列{bn},b1=2,5S5=2S8
==>5(5b1+10d)=2(8b1+28d)
==>d=3, ∴bn=3n-1
Tn=2×1+5×4+8×16+...+(3n-1)×4^(n-1) (1)
4Tn=2×4+5×16+......+(3n-4)×4^(n-1)+(3n-1)×4^n (2)
(1)-(2): -3Tn=2+3[4+16+......+4^(n-1)]-(3n-1)×4^n
=2+12[4^(n-1)-1]/3-(3n-1)×4^n
=-2 -(3n-2)×4^n
Tn=2/3+(n-2/3)×4^n
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