Question

已知数列an 的前n项和为Sn，且Sn=2an-2（n=1，2，3，…）

已知数列{an}的前n项和为Sn,且Sn=2an-2(n=1,2,3……),数列{bn}中b1=1,点P(bn,b(n+1)在直线X-Y+2=0上(1)求数列{an},{bn}的通项an,bn(2)设Cn=an*bn,求数列{Cn}的前N项和Tn,并求满足Tn<167的最大正整数n

关键讲第二问的错位相减法，减出来是什么。结果？

Answer 1

s1=a1=2a1-2
=>a1=2
s(n+1)=2a(n+1)-2=sn+a(n+1)=2sn-2+a(n+1)
=>a(n+1)=2an
=>an=2^n
x-y+2=0
=>y=x+2
=>(bn,b(n+1))满足关系b(n+1)=bn+2
b1=1
=>bn=2n-1
cn=an*bn
=>cn=(2n-1)*(2^n)
=>2Cn=(2n-1)*(2^(n+1))
Tn=Cn+C(n-1)……+C2+C1=(2n-1)*2^n+(2*(n-1)-1)*2^(n-1)+……+12+2
2*Tn=(2n-1)*2^(n+1)+(2*(n-1)-1)*2^n+……+24+4=(2n-1)*2^(n+1)+(2*n-3)*2^n+……+24+4
2*Tn-Tn=(2n-1)*2^(n+1)-2*2^n-……-16-8-2=(2n-1)*2^(n+1)-8*(2^n-1)+8-2=(2n-3)*2^(n+1)+6
Tn<167
=>(2n-3)*2^(n+1)+6<167
=>n=5