Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)+1\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n^2+2n+n+2}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n^2+3n+2}\right)\)
\(S_n=\dfrac{1}{4}-\dfrac{1}{2\left(n^2+3n+2\right)}\)
\(S_n=\dfrac{1}{4}-\dfrac{1}{2n^2+6n+4}\)
\(S_n=\dfrac{2n^2+6n+4}{4\left(2n^2+6n+4\right)}-\dfrac{4}{4\left(2n^2+6n+4\right)}\)
\(S_n=\dfrac{2n^2+6n+4}{8n^2+48n+16}-\dfrac{4}{8n^2+48n+16}\)
\(S_n=\dfrac{2n^2+6n}{8n^2+48n+16}\)
\(S_n=\dfrac{2\left(n^2+3n\right)}{2\left(4n^2+24n+8\right)}=\dfrac{n^2+3n}{4n^2+24n+8}\)
\(S_n=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\\ 2S_n=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\\ 2S_n=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\\ =\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\\ =\dfrac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\\ =>S_n=\dfrac{\left(n+1\right)\left(n+2\right)-2}{4\left(n+1\right)\left(n+2\right)}\)
Giải sai r nhéLinh Nguyễn
Lời giải:
a) \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{n-1}-1\right)\left(\frac{1}{n}-1\right)\)
\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}...\frac{-(n-2)}{n-1}.\frac{-(n-1)}{n}\)
\(=\frac{(-1)(-2)(-3)...[-(n-2)][-(n-1)]}{2.3.4...(n-1)n}\)
\(=\frac{(-1)^{n-1}(1.2.3....(n-2)(n-1))}{2.3.4...(n-1)n}=(-1)^{n-1}.\frac{1}{n}\)
b) \(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{n^2}-1\right)\)
\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.....\frac{1-n^2}{n^2}\)
\(=\frac{(-1)(2^2-1)}{2^2}.\frac{(-1)(3^2-1)}{3^2}....\frac{(-1)(n^2-1)}{n^2}\)
\(=(-1)^{n-1}.\frac{(2^2-1)(3^2-1)...(n^2-1)}{2^2.3^2....n^2}\)
\(=(-1)^{n-1}.\frac{(2-1)(2+1)(3-1)(3+1)...(n-1)(n+1)}{2^2.3^2....n^2}\)
\(=(-1)^{n-1}.\frac{(2-1)(3-1)...(n-1)}{2.3...n}.\frac{(2+1)(3+1)...(n+1)}{2.3...n}\)
\(=(-1)^{n-1}.\frac{1.2.3...(n-1)}{2.3...n}.\frac{3.4...(n+1)}{2.3.4...n}\)
\(=(-1)^{n-1}.\frac{1}{n}.\frac{n+1}{2}=(-1)^{n-1}.\frac{n+1}{2n}\)
a) \(\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^{n-1}}\)
\(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n:\left(-\dfrac{5}{7}\right)}\)
\(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n.\left(-\dfrac{7}{5}\right)}\)
\(=\dfrac{1}{\left(-\dfrac{7}{5}\right)}\)
\(=1.\left(-\dfrac{5}{7}\right)\)
\(=-\dfrac{5}{7}\)
b) \(\dfrac{\left(-\dfrac{1}{2}\right)^{2n}}{\left(-\dfrac{1}{2}\right)^n}\)
\(=\dfrac{\left(-\dfrac{1}{2}\right)^n.\left(-\dfrac{1}{2}\right)^n}{\left(-\dfrac{1}{2}\right)^n}\)
\(=\left(-\dfrac{1}{2}\right)^n\)
\(N=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\Rightarrow2N=2+1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow N=2N-N=2+1+\dfrac{1}{2}+...+\left(\dfrac{1}{2}\right)^{99}-1-\dfrac{1}{2}-...-\left(\dfrac{1}{2}\right)^{100}=2-\left(\dfrac{1}{2}\right)^{100}\)
\(N=1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\)
\(\dfrac{1}{2}N-N=\left(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\right)\)
\(-\left(1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\right)\)
\(-\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}-1\)
\(N=\dfrac{-\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}}{-\dfrac{1}{2}}\)
a: \(=\left(-\dfrac{5}{7}\right)^{n-n}=\left(-\dfrac{5}{7}\right)^0=1\)
b: \(=\left(-\dfrac{1}{2}\right)^{2n-n}=\left(-\dfrac{1}{2}\right)^n\)
\(S_n=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+....+\dfrac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(S_n=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+....+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}-\dfrac{1}{n\left(n+2\right)\left(n+3\right)}\right)\)\(S_n=\dfrac{1}{3}\left(\dfrac{1}{2.3.4}-\dfrac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right)\)
\(S_n=\dfrac{1}{3}\left(\dfrac{1}{24}-\dfrac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right)\)
\(S_n=\dfrac{1}{72}-\dfrac{1}{3\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)