D = $\frac{2}{3}.\frac{5}{6}.\frac{9}{10}. ... .\frac{799}{780}$

   = $\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}. ... .\frac{38.41}{39.40}$

   = $\frac{2.2}{2.3}.\frac{2.3. ... .38}{3.4. ... 39}.\frac{5.6. ... .41}{4.5. ... .40}$

   = $\frac{2}{3}.\frac{2}{39}.\frac{41}{4}$

   = $\frac{41}{3.39}$

D = \(\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}.....\frac{779}{780}\)

   = \(\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}.....\frac{38.41}{39.40}\)

   = \(\frac{2}{3}.\frac{2.3.4....38}{3.4.5....39}.\frac{5.6.7.....41}{4.5.6.....40}\)

   = \(\frac{2}{3}.\frac{2}{39}.\frac{41}{4}\)

   = \(\frac{41}{117}\)

Ở tổng A bạn có chép sai đề không vậy?

\(D=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right)\)

\(D=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{779}{780}\)

\(D=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{1558}{1560}\)

\(D=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...\frac{38.41}{39.40}\)

\(D=\frac{1.2.3.4...38}{3.4.5.6...40}.\frac{4.5.6.7...41}{2.3.4.5...39}\)

\(D=\frac{2}{39.40}.\frac{40.41}{2.3}\)

\(D=\frac{41}{39.3}=\frac{41}{117}\)

A=(−2/3 ).(−5/6 ).(−9/10 )...(−779/780 )=(−4/6 ).(−10/12 ).(−18/20 )....(−1558/1560 )

A=(−1.4).(−2.5).(−3.6)....(−38.41) / (2.3).(3.4).(4.5)....(39.40) =(1.2.3....38).(4.5.6...41) / (2.3.4...39).(3.4.5...40)  ( Vì từ -1 đến -38 có 38 số =>tích của 38 số âm = tích của 38 số dương)

A=(1.2.3....38).(4.5.6...41) / (2.3.4...39).(3.4.5...40) =1.41/39.3 =41/ 117 

D=2/3.5/6.9/10.14/15...779/780

D=4/6.10/12.18/20.28/30...1558/1560

D=1.4/2.3   .    2.5/3.4    .     3.6/4.5     .      4.7/5.6         ....      38.41/39.40

D=(1.2.3....38).(4.5.6...41)/(2.3.4...39).(3.4.5...40)

D=1.41/39.3

D= 41/117

Xét trường hợp n chẵn:

\(1^2+2^2+3^2+...+n^2=\left(1^2+3^2+5^2+...+\left(n-1\right)^2\right)+\left(2^2+4^2+6^2+...+n^2\right)\)

\(=\frac{\left(n-1\right).n.\left(n+1\right)+n\left(n+1\right).\left(n+2\right)}{6}\)

\(=\frac{n\left(n+1\right).\left(n-1+n+2\right)}{6}\)

\(=\frac{n\left(n+1\right).\left(2n+1\right)}{6}\)

Tương tự với trường hợp n lẻ . ta có \(\text{ĐPCM}\)

\(A=1^2+2^2+3^2+....+n^2\)

\(=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+....+n\left[\left(n+1\right)-1\right]\)

\(=1.2-1+2.3-2+3.4-3+...+n\left(n+1\right)-n\)

\(=\left[1.2+2.3+3.4+....+n\left(n+1\right)\right]-\left(1+2+3+....+n\right)\)

Ta có :

\(1.2+2.3+3.4+....+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)(cái này tự CM nha)

\(1+2+3+....+n=\frac{n\left(n+1\right)}{2}\)

\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)(đpcm)

Cảm ơn 2 bạn nha

\(=\frac{3.8.15........\left(n-1\right)\left(n+1\right)}{\left(2.3.4.....n\right)\left(2.3.4......n\right)}=\frac{1.3.2.4.3.5..............\left(n-1\right)\left(n+1\right)}{\left(2.3.4.....n\right)\left(2.3.4......n\right)}=\frac{\left(1.2.3......\left(n-1\right)\right)\left(3.4.5......\left(n+1\right)\right)}{\left(2.3.4....n\right)\left(2.3.4.......n\right)}\)

\(=\frac{1.\left(n+1\right)}{n.2}=\frac{n+1}{2n}\)

Bài này mình làm rồi còn gì?

Ta có: \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)....\left(1-\frac{1}{780}\right)\)

\(=\frac{2}{3}.\frac{5}{6}...\frac{779}{780}\)

\(=\frac{4}{6}.\frac{10}{12}....\frac{1558}{1560}\)

\(=\frac{1.4.2.5....38.41}{2.3.3.4....39.40}=\frac{\left(1.2.3..38\right)\left(4.5...41\right)}{\left(2.3.4...39\right)\left(3...40\right)}=\frac{41}{39.3}=\frac{41}{117}\)

 \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)........\left(1-\frac{1}{780}\right)\)

\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}........\frac{779}{780}\)

\(=\frac{4}{6}.\frac{10}{12}\frac{18}{20}.\frac{28}{30}.........\frac{1558}{1560}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...............\frac{38.41}{39.40}\)

\(=\frac{\left(1.2.3.4......38\right)\left(4.5.6.7..........41\right)}{\left(2.3.4.5.........39\right)\left(3.4.5.6.........40\right)}\)

\(=\frac{1.41}{39.3}\)

\(=\frac{41}{117}\)

Vậy \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)........\left(1-\frac{1}{780}\right)=\frac{41}{117}\)

\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right)\)

\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{779}{780}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{1558}{1560}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{38.41}{39.40}\)

\(=\frac{\left(1.2.3...38\right)\left(4.5.6...61\right)}{\left(2.3.4...39\right)\left(3.4.5...40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)

\(\left(1-\frac{1}{3}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{10}\right).\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right).a=1\)

\(\left(\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{779}{780}\right).a=1\)

\(\left(\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{1558}{1560}\right).a=1\)

\(\left(\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...\frac{38.41}{39.40}\right).a=1\)

\(\left(\frac{1.2.3.4...38}{3.4.5.6..40}.\frac{4.5.6.7...41}{2.3.4.5..39}\right).a=1\)

\(\left(\frac{2}{39.40}.\frac{40.41}{2.3}\right).a=1\)

\(\frac{41}{39.3}.a=1\)

\(\frac{41}{117}.a=1\)

\(a=1:\frac{41}{117}\)

\(a=1.\frac{117}{41}=\frac{117}{41}\)

Vậy a = 117/41

Ủng hộ mk nha ^_-

các bn giups mk đi mai mk phải nộp bài rùi

\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right)\)

\(\Rightarrow B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{779}{780}\)

\(\Rightarrow B=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{1558}{1560}\)

\(\Rightarrow B=\frac{1.4}{2.3}.\frac{2.5}{3.4}\frac{3.6}{4.5}...\frac{38.41}{39.40}\)

\(\Rightarrow B=\frac{\left(1.2.3...38\right)\left(4.5.6...41\right)}{\left(2.3.4...39\right)\left(3.4.5...40\right)}\)

\(\Rightarrow B=\frac{1.41}{39.3}=\frac{41}{117}\)

Vậy B=\(\frac{41}{117}\)

Ai thấy đúng thì k nha

B= \(\frac{41}{117}\)