问题
填空题
已知数列{an}满足a1=1,an+an+1=(
|
答案
由Sn=a1+3•a2+32•a3+…+3n-1•an,
所以3Sn=3a1+32•a2+33•a3+…+3n•an,
相加4Sn=a1+3(a1+a2)+…+3n-1•(an-1+an)+3n•an,
所以4Sn-3nan=1+3(
)1+32(1 3
)2+…+3n-1•(1 3
)n-1=n.1 3
故答案为n.
已知数列{an}满足a1=1,an+an+1=(
|
由Sn=a1+3•a2+32•a3+…+3n-1•an,
所以3Sn=3a1+32•a2+33•a3+…+3n•an,
相加4Sn=a1+3(a1+a2)+…+3n-1•(an-1+an)+3n•an,
所以4Sn-3nan=1+3(
)1+32(1 3
)2+…+3n-1•(1 3
)n-1=n.1 3
故答案为n.