已知等比數列{an},a1>0,公比q>-1,且不等于0。bn=a(n+1)+a(n+2)，n屬于正整數。{an},{bn}的前n項和分別為An,Bn，試比較An,Bn的大小

熱心網友

解:1)q=1時,An=ｎa1,Bn=2ｎa1,∵a10∴An＜Bn2)q≠1時,An=[a1-a(n+1)]/(1-q)∵bn=a(n+1)+a(n+2)，b1=a2+a3，∴Bn=[a2-a(n+2)]/(1-q)+[a3-a(n+3)]/(1-q)∴(1-q)(An-Bn)=(a1-a2-a3)-{a(n+1)-a(n+2)-a(n-3)]=(a1-a2-a3)-(a1-a2-a3)q^n=a1(1-q-q^)(1-q^n)∴An-Bn=a1(1-q-q^)[(1-q^n)/(1-q)]下面證明:[(1-q^n)/(1-q)]＞0當-1＜q＜1時:1-q＞0且1-q^n＞0∴[(1-q^n)/(1-q)]＞0當q＞1時:1-q＜0且1-q^n＜0∴[(1-q^n)/(1-q)]＞0由于[(1-q^n)/(1-q)]＞0且a1＞0∴An-Bn=a1(1-q-q^)[(1-q^n)/(1-q)]關鍵是看(1-q-q^)與0的大小令1-q-q^=0,q=(-1+√5)/2或q=(-1-√5)/2q=(-1-√5)/2＜-1不符合題意舍去,∴當-1＜q＜(-1+√5)/2時:An＞Bn。當q=(-1+√5)/2時:An=Bn。當q＞(-1+√5)/2時:An＜Bn。綜合1),2)可得:當-1＜q＜(-1+√5)/2時:An＞Bn,當q=(-1+√5)/2時:An=Bn,當q＞(-1+√5)/2時:An＜Bn。