• Giả sử $d>0$. Ta có:
$d= u_{2} - u_{1} =..........= u_{n}-u_{n-1}$
Do đó :
$\frac{ 1}{ u_{1} u_{2} }= \frac{ u_{2} – u_{1} }{ u_{1} u_{2} }. \frac{ 1}{d}= \frac{ 1}{d} \left( \frac{ 1}{u_{1}}-\frac{ 1}{u_{2}} \right) $
$\frac{ 1}{ u_{2} u_{3} }= \frac{ 1}{d} \left( \frac{ 1}{u_{2}}-\frac{ 1}{u_{3}} \right) $
$\frac{ 1}{ u_{3} u_{4} }= \frac{ 1}{d} \left( \frac{ 1}{u_{3}}-\frac{ 1}{u_{4}} \right) $
………
$\frac{ 1}{ u_{n-1} u_{n}} =\frac{ 1}{d} \left( \frac{ 1}{u_{n-1}}-\frac{ 1}{u_{n}} \right) $
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$\Rightarrow S=\frac{ 1}{d} \left(\frac{ 1}{ u_{1} }-\frac{ 1}{u_{n}} \right) = \frac{ 1}{d}.\frac{ u_{n}-u_{1} }{ u_{1} u_{n}}= \frac{ 1}{d}. \frac{ u_{1} + \left( n-1 \right)d- u_{1} }{ u_{1} u_{n}} $
$\Rightarrow S=\frac{n-1}{u_1u_n}.$
• $d<0$ chứng minh tương tự.
Vậy ta có đpcm.