已知{an}是等差数列,其前n项和为sn,{bn}是等比数列,且a1=b1=2,a4+b4=27,s4-b4=10
已知{an}是等差数列,其前n项和为sn,{bn}是等比数列,且a1=b1=2,a4+b4=27,s4-b4=10
(1) 求数列{an}与{bn}的通项公式
(2)记Tn=anb1+an-1b2+...+a1bn,证明Tn+12=-2an+10bn (n∈N+)
最好可以是写在纸上拍下来的答案,方便思维嘛.
只求第二问
Tn=2an+22an-1+23an-2+…+2na1; ①;
2Tn=22an+23an-1+…+2na2+2n+1a1; ②;
由②-①得,Tn=-2(3n-1)+3×22+3×23+…+3×2n+2n+2
=
12(1-2 n-1)
1-2
+2n+2-6n+2
=10×2n-6n-10;
而-2an+10bn-12=-2(3n-1)+10×2n-12=10×2n-6n-10;
故Tn+12=-2an+10bn(n∈N*).写错了吧??? 不好意思啊没看懂。能写在纸上吗,会加分的不是啊,之前我没有想到你会看不懂我再打清楚些,照片我拍了可上传时间太久了。同学我再写清楚一些,你试着看懂好吗~(1)An=3n-1 Bn=2^n这个第一小题不解释了啊(2)Tn=(3n-1)*2+(3n-4)*2^2+(3n-7)*2^3+......+8*2^(n-2)+5*2^(n-1)+2*2^n① 2Tn=(3n-1)*2^2+(3n-4)*2^3+(3n-7)*2^4+......+8*2^(n-1)+5*2^n+2*2^(n+1)②①-②,得Tn=-(3n-1)*2 + 3[2^2 + 2^3 + ... + 2^n] + 2^(n+2) =2^(n+2) - 2(3n-1) + 12[1+2+...+2^(n-2)] =2^(n+2)-2(3n-1)+12[2^(n-1)-1] =2*2^(n+1)-6n+2 +3*2^(n+1)-12 =5*2^(n+1) - 6n - 10PS:“^”是指乘方-2An+10Bn=-2(3n-1)+10*2^n=5*2^(n+1)-6n+2=Tn+12