a) Giả sử $\div (u_n)$ có công sai $d\neq 0 $, ta có:
$\dfrac{S_m}{S_n}= \dfrac{\left[ {(2u_1+(m-1)d)} \right]m }{\left[ {2u_1+(n-1)d} \right]n }=\dfrac{m^2}{n^2} \Rightarrow \dfrac{2u_1+(m-1)d}{2u_1+(n-1)d}=\dfrac{m}{n}$
hay $2u_1.n+(m-1)nd=2u_1.m+(n-1)m.d$
$\Leftrightarrow 2u_1(m-n)-(m-n)d=0 $
$\Rightarrow d=2u_1$ vì $m\neq n$
Do đó $\dfrac{u_m}{u_n}=\dfrac{u_1+(m-1)d}{u_1+(n-1)d}=\dfrac{u_1(1+2(m-1))}{u_1(1+2(n-1))}=\dfrac{2m-1}{2n-1} $ (đpcm).
b) $\dfrac{S_m}{S_n}=1 \Leftrightarrow S_m=S_n \Leftrightarrow \left[ {2u_1+(m-1)d} \right]m=\left[ {2u_1+(n-1)d} \right]n $
$\Leftrightarrow 2u_1m+m(m-1)d=2u_1n+n(n-1)d $
$\Leftrightarrow 2u_1(m-n)+(m^2-n^2)d-(m-n)d=0 (1)$
Vì $m\neq n$, từ $(1)$ suy ra : $2u_1+(m+n-1)d=0$
Do đó: $S_{m+n}=\dfrac{ \left[ {2u_1+(m+n-1)s} \right](m+n) }{2} =0$ (đpcm)