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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.
\(B=\left(\frac{3}{5}\right)^2\cdot5^2-\left(2\frac{1}{4}\right)^3:\left(\frac{3}{4}\right)^3+\frac{1}{2}\)
\(B=\left(\frac{3}{5}\cdot5\right)^2-\left(\frac{9}{4}:\frac{3}{4}\right)^3+\frac{1}{2}\)
\(B=3^2-\left(\frac{9}{4}\cdot\frac{4}{3}\right)^3+\frac{1}{2}\)
\(B=3^2-3^3+\frac{1}{2}=-18+\frac{1}{2}=-\frac{35}{2}\)
\(A=\left(1+\frac{1}{3}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n^2+2n}\right)\)
\(A=\frac{3+1}{3}.\frac{8+1}{8}.\frac{15+1}{15}...\frac{n^2+2n+1}{n^2+2n}\)
\(A=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{\left(n+1\right)^2}{n^2+2n}\)
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(A=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)
\(A=\left(n+1\right).\frac{2}{n+2}=\frac{2.\left(n+1\right)}{n+2}\)
Ta có : \(1+\frac{1}{k^2+2k}=\frac{k^2+2k+1}{k^2+2k}=\frac{\left(k+1\right)^2}{k\left(k+2\right)}\) với k thuộc N*
Áp dụng với k = 1,2,3,....,n được :
\(A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n^2+2n}\right)\)
\(=\frac{\left(1+1\right)^2}{1.\left(1+2\right)}.\frac{\left(2+1\right)^2}{2.\left(2+2\right)}.\frac{\left(3+1\right)^2}{3.\left(3+2\right)}...\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(=\frac{\left[2.3.4...\left(n+1\right)\right]^2}{1.2.3...n.3.4.5...\left(n+2\right)}=\frac{\left[\left(n+1\right)!\right]^2}{n!.\frac{\left(n+2\right)!}{2}}\)
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!
