1/1x2x3+1/2x3x4+...1/118x19x20<1/4 <--- cái này đề sai ở 1/118x19x20 phải là 1/18x19x20

Ta có:

\(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)

\(=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{5n+6}\right)=\frac{1}{5}\left(\frac{5n+6}{5n+6}-\frac{1}{5n+6}\right)=\frac{1}{5}.\frac{5n+5}{5n+6}=\frac{1}{5}.\frac{5\left(n+1\right)}{5n+6}=\frac{5\left(n+1\right)}{5\left(5n+6\right)}=\frac{n+1}{5n+6}\)(ĐPCM)

bạn Phạm Thiết Tường ơi ch mình hỏi sao lại nhân \(\frac{1}{5}\)với \(\frac{1}{1}-\frac{1}{5n+6}\)vậy

@tranlethanhtruc: Bn ấy lm sai.

đấy có phải lớp 4 ko đấy

Lớp 4 đó nếu ai ko làm được thì ko phải học sinh giỏi đó nha

bạn lấy đâu ra cái bài này vậy 

hơi căng đấy

 CM:  \(\dfrac{1}{1.6}\)+ \(\dfrac{1}{11.16}\)+...+ \(\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\) = \(\dfrac{n+1}{5n+6}\)

A = \(\dfrac{1}{5}\)(\(\dfrac{5}{1.6}\) + \(\dfrac{5}{6.11}\)+...+ \(\dfrac{5}{\left(5n+1\right).\left(5n+6\right)}\)) 

A = \(\dfrac{1}{5}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{6}\)+ \(\dfrac{1}{6}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{5n+1}\) - \(\dfrac{1}{5n+6}\))

A = \(\dfrac{1}{5}\) .( \(\dfrac{1}{1}\) - \(\dfrac{1}{5n+6}\))

A = \(\dfrac{1}{5}\). \(\dfrac{5n+6-1}{5n+6}\)

A = \(\dfrac{1}{5}\). \(\dfrac{5n+5}{5n+6}\)

A = \(\dfrac{1}{5}\) . \(\dfrac{5.\left(n+1\right)}{5n+6}\)

A = \(\dfrac{n+1}{5n+6}\)

⇒\(\dfrac{1}{1.6}\) + \(\dfrac{1}{6.11}\)+ \(\dfrac{1}{11.16}\)+...+ \(\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\) = \(\dfrac{n+1}{5n+1}\) (đpcm)

\(A=\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\)

\(A=\dfrac{1}{5}\left[1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right]\)

\(A=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)

\(A=\dfrac{1}{5}\left(\dfrac{5n+6-1}{5n+6}\right)=\dfrac{1}{5}\left(\dfrac{5n+5}{5n+6}\right)=\dfrac{1}{5}.5\left(\dfrac{n+1}{5n+6}\right)=\dfrac{n+1}{5n+6}\)

\(\Rightarrow dpcm\)

Ta có:\(\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)

        \(=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)

        \(=\frac{1}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)

        \(=\frac{1}{5}.\left(1-\frac{1}{5n+6}\right)\)

        \(=\frac{1}{5}.\left(\frac{5n+5}{5n+6}\right)=\frac{n+1}{5n+6}\left(\text{đ}pcm\right)\)

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