数列{an}满足a1=1,a2=3/2,a(n+2)=3/2a(n+1)-1/2an(n∈N*)(1)记dn=a(n+1)-an,求证:{dn}是等比数列(2)求数列{an}的通向公式
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数列{an}满足a1=1,a2=3/2,a(n+2)=3/2a(n+1)-1/2an(n∈N*)(1)记dn=a(n+1)-an,求证:{dn}是等比数列(2)求数列{an}的通向公式
数列{an}满足a1=1,a2=3/2,a(n+2)=3/2a(n+1)-1/2an(n∈N*)
(1)记dn=a(n+1)-an,求证:{dn}是等比数列
(2)求数列{an}的通向公式
数列{an}满足a1=1,a2=3/2,a(n+2)=3/2a(n+1)-1/2an(n∈N*)(1)记dn=a(n+1)-an,求证:{dn}是等比数列(2)求数列{an}的通向公式
(1)由a(n+2)=3/2a(n+1)-1/2an 得, 2*a(n+2)=3*a(n+1)-an , 2*[a(n+2)-a(n+1)]=a(n+1)-an
即 2d(n+1)=dn , d(n+1)/dn=1/2
所以 :{dn}是等比数列
(2) d1=a2-a1=1/2 ,{dn}的前n项和为 S1=1-(1/2)^n
S1= d1+d2+……+dn=(a2-a1)+(a3-a2)+……+[a(n+1)-an]=a(n+1)-a1=a(n+1)-1=1-(1/2)^n
a(n+1)=2-(1/2)^n ,an=2-(1/2)^(n-1)
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