Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A= \(\left(\frac{3}{4}\right)\left(\frac{8}{9}\right)\left(\frac{15}{16}\right)......\left(\frac{\left(n-1\right)\left(n+1\right)}{n.n}\right)\)
\(=\frac{3.8.15....\left(n-1\right)\left(n+1\right)}{\left(2.3.4......n\right)\left(2.3.4.......n\right)}=\frac{1.3.2.4.3.5.......\left(n-1\right)\left(n+1\right)}{\left(2.3.4.....n\right)\left(2.3.4..................n\right)}=\frac{\left(1.2.3.......\left(n-1\right)\right)\left(3.4.5........\left(n+1\right)\right)}{\left(2.3.4.....n\right)\left(2.3.4...........n\right)}\)
\(=\frac{1.\left(n+1\right)}{n.2}=\frac{n+1}{2n}\)
mình chỉ tick cho những người giải thôi, không chấp nhận trường hợp xin tick, và cấm tình trạng spam bậy. Nếu ai giải được thì mình tick, nếu ai không giải, xin tick, hay spam để kiếm điểm hỏi đáp thì miễn.
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n}{n+1}\)
\(=\frac{1}{n+1}\)
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)...+\frac{1}{20}.\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+\frac{1}{4}.4.5:2+...+\frac{1}{20}.20.21:2\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+5+...+21}{2}=115\)
Ta có:
\(\left(1-\dfrac{1}{2^2}\right).\left(1-\dfrac{1}{3^2}\right).\left(1-\dfrac{1}{4^2}\right).....\left(1-\dfrac{1}{n^2}\right)\)
\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.....\dfrac{n^2-1}{n^2}\)
\(=\dfrac{3.8.15....\left(n^2-1\right)}{4.9.16.....n^2}\)
\(=\dfrac{1.3.2.4.3.5....\left(n-1\right)\left(n+1\right)}{2.2.3.3.4.4....n.n}\)
\(=\dfrac{\left[1.2.3....\left(n-1\right)\right].\left[3.4.5....\left(n+1\right)\right]}{\left(2.3.4....n\right).\left(2.3.4....n\right)}\)
\(=\dfrac{1.\left(n+1\right)}{n.2}=\dfrac{n+1}{2n}\)
Ta có công thức:
\(1-\dfrac{1}{k^2}=\dfrac{k^2-1^2}{k^2}=\dfrac{\left(k+1\right)\left(k+2\right)}{k^2}\)
Áp dụng công thức trên ta đc:
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)....\left(1-\dfrac{1}{n^2}\right)\)
\(=\dfrac{2^2-1^2}{2^2}.\dfrac{3^2-1^2}{3^2}.\dfrac{4^2-1^2}{4^2}....\dfrac{n^2-1^2}{n^2}\)
\(=\dfrac{\left(2+1\right)\left(2-1\right)}{2.2}.\dfrac{\left(3+1\right)\left(3-1\right)}{3.3}.\dfrac{\left(4+1\right)\left(4-1\right)}{4.4}....\dfrac{\left(n+1\right)\left(n-1\right)}{n.n}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}....\dfrac{\left(n+1\right)\left(n-1\right)}{n.n}\)
\(=\dfrac{[1.2.3....\left(n+1\right)].[3.4.5....\left(n-1\right)]}{\left(2.3.4....n\right)\left(2.3.4....n\right)}\)
\(=\left(n+1\right).\dfrac{1}{2n}=\dfrac{n+1}{2n}\)
Chúc bạn học tốt!
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)
\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2001}{2002}\right)\left(-\frac{2002}{2003}\right)\)
\(=\frac{-1.\left(-2\right).....\left(-2001\right)\left(-2002\right)}{2.3....2002.2003}\)
\(=\frac{1}{2003}\)
\(=\frac{3.8.15........\left(n-1\right)\left(n+1\right)}{\left(2.3.4.....n\right)\left(2.3.4......n\right)}=\frac{1.3.2.4.3.5..............\left(n-1\right)\left(n+1\right)}{\left(2.3.4.....n\right)\left(2.3.4......n\right)}=\frac{\left(1.2.3......\left(n-1\right)\right)\left(3.4.5......\left(n+1\right)\right)}{\left(2.3.4....n\right)\left(2.3.4.......n\right)}\)
\(=\frac{1.\left(n+1\right)}{n.2}=\frac{n+1}{2n}\)
Bài này mình làm rồi còn gì?