Question

Chứng tỏ:

\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}=\dfrac{1}{4}\)

(Lưu ý dấu "." trong phân số là dấu nhân)

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#Toán lớp 6     

Answer 1

\(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + .....+ \(\dfrac{1}{10.11.12}\)

= \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}\) + \(\dfrac{1}{2.3}\) - \(\dfrac{1}{3.4}\) +....+ \(\dfrac{1}{10.11}\) - \(\dfrac{1}{11.12}\)

=\(\dfrac{1}{1.2}\) + (- \(\dfrac{1}{2.3}\) + \(\dfrac{1}{2.3}\))+.......+ ( \(-\dfrac{1}{10.11}\) + \(\dfrac{1}{10.11}\)) - \(\dfrac{1}{11.12}\)

=\(\dfrac{1}{2}\) - \(\dfrac{1}{11.12}\) =\(\dfrac{1}{2}\) - \(\dfrac{1}{132}\) =\(\dfrac{66}{132}\)-\(\dfrac{1}{132}\) =\(\dfrac{65}{132}\) Vì \(\dfrac{33}{132}\) = \(\dfrac{1}{4}\) nên \(\dfrac{65}{132}\) > \(\dfrac{1}{4}\)