Ta chứng minh $Q_{n+1}=P_n(u_{n+1}-1)+au_{n+1}=0$
$\Leftrightarrow P_n(\frac{u_n^2}{u_n^2-u_n+1}-1)+a.\frac{u_n^2}{u_n^2-u_n+1}=0$
$\Leftrightarrow P_n\frac{u_n-1}{u_n^2-u_n+1}+\frac{au_n^2}{u_n^2-u_n+1}=0$
$\Leftrightarrow P_n(u_n-1)+au_n^2=0$
$\Leftrightarrow P_{n-1}.u_n.(u_n-1)+au_n^2=0$
$\Leftrightarrow P_{n-1}(u_n-1)+au_n=0$
$\Leftrightarrow Q_n=0$
$\Leftrightarrow ....$
$\Leftrightarrow Q_0=0$ Đúng (Quy ước $P_0=1)$
Quay lại với bài toán
$T_n=aS_n+P_n$
$T_{n+1}-T_n=a(S_{n+1}-S_n)+P_{n+1}-P_n$
$=au_{n+1}+P_n(u_{n+1}-1)$
$=Q_{n+1}$
$=0$
$\Rightarrow T_{n+1}=T_n=....=T_1=1$