Ta chứng minh Qn+1=Pn(un+1−1)+aun+1=0
⇔Pn(u2nu2n−un+1−1)+a.u2nu2n−un+1=0
⇔Pnun−1u2n−un+1+au2nu2n−un+1=0
⇔Pn(un−1)+au2n=0
⇔Pn−1.un.(un−1)+au2n=0
⇔Pn−1(un−1)+aun=0
⇔Qn=0
⇔....
⇔Q0=0 Đúng (Quy ước P0=1)
Quay lại với bài toán
Tn=aSn+Pn
Tn+1−Tn=a(Sn+1−Sn)+Pn+1−Pn
=aun+1+Pn(un+1−1)
=Qn+1
=0
⇒Tn+1=Tn=....=T1=1