Question

等比數(shù)列{an}的首項a1=，且4an-1+an+1=4an,則sina1+sina2+sina3+…+sina2014=

 

Answer 1

+

【解析】設(shè)等比數(shù)列的公比為q

∵4an-1+an+1=4an

∴4an-1+an-1q2=4an-1q

∴4+q2=4qq=2an=·2n-1

∴a1=, a2=·2=, a3=·22=,a4=·23=,a5=·24=, a6=·24=,…

∴sina1=sin=,sina2=sin=,

sina3=sin= sin()=,

sina4=sin(·22)=sin=sin()=-sin=-

sina5=sin(·23)=sin=sin()=sin= sin()=

sina6=sin(·24)=sin= sin()=sin()=-sin=-

一般地,當(dāng)n≥3時,設(shè)n=2k+1(k=1,2,3,…),則

an=·2n-1=·22k-2=·4k-1=·(1+3)k-1=·(1+3+32+…+3k-1)

=+(+31+…+3k-2)(規(guī)定=0)

∴sinan=sin[+(+31+…+3k-2)]=sin= sin()=,

設(shè)n=2k+2(k=1,2,3,…),則

an=·2n-1=·22k-2=·4k-1=·(1+3)k-1=·(1+3+32+…+3k-1)

=+(+31+…+3k-2)

∴sinan=sin[+(+31+…+3k-2)]= sin=sin()=-sin=-

所以,當(dāng)n≥3時,sina2k+1+sina2k+2=0(k=1,2,3,…)

∴sina1+sina2+sina3+…+sina2014= sina1+sina2=+