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a) \(1.2+2.3+...+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)(@@)
+) Với n = 1 ta có: \(1.2=\frac{1.\left(1+1\right)\left(1+2\right)}{3}\) đúng
=> (@@) đúng với n = 1
+) G/s (@@) đúng cho đến n
+) Ta chứng minh (@@ ) đúng với n + 1
Ta có: \(1.2+2.3+...+n\left(n+1\right)+\left(n+1\right)\left(n+2\right)\)
\(=\frac{n\left(n+1\right)\left(n+2\right)}{3}+\left(n+1\right)\left(n+2\right)\)
\(=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right)}{3}\)
=> (@@) đúng với n + 1
Vậy (@@ ) đúng với mọi số tự nhiên n khác 0
b) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}=\frac{2^n-1}{2^n}\) (@)
Ta chứng minh (@) đúng với n là số tự nhiên khác 0 quy nạp theo n
+) Với n = 1 ta có: \(\frac{1}{2}=\frac{2^1-1}{2^1}\) đúng
=> (@) đúng với n = 1
+) G/s (@) đúng cho đến n
+) Ta cần chứng minh (@) đúng với n + 1
Ta có: \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}+\frac{1}{2^{n+1}}=\frac{2^n-1}{2^n}+\frac{1}{2^{n+1}}=\frac{2^{n+1}-2+1}{2^{n+1}}=\frac{2^{n+1}-1}{2^{n+1}}\)
=> (@) đúng với n + 1
Vậy (@) đúng với mọi số tự nhiên n khác 0.
CM : \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
Có : \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}\)\(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\) đpcm
\(\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{1}{2}\left(\frac{\left(n+1\right)\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}-\frac{n\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}\right)\)
\(\frac{1}{2}\left(\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}-\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{1}{2}\left(\frac{\left(n+1\right)\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}-\frac{n\left(n+1\right)}{n\left(n+1\right)\left(n+1\right)\left(n+2\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}-\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+2\right)\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\frac{3n}{2.\left(3n+2\right)}\)
\(=\frac{n}{2\left(3n+2\right)}\)
a) Nhân cả tử và mẫu với 2 . 4 . 6 ... 40 ta được :
\(\frac{1.3.5...39}{21.22.23...40}=\frac{\left(1.3.5...39\right).\left(2.4.6...40\right)}{\left(21.22.23...40\right).\left(2.4.6...40\right)}\)
\(=\frac{1.2.3...39.40}{1.2.3...40.2^{20}}=\frac{1}{2^{20}}\)
b) Nhân cả tử và mẫu với 2 . 4 . 6 ... 2n ta được :
\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3....2n\right)}=\frac{1.3.5...\left(2n-1\right).\left(2.4.6...2n\right)}{\left(n+1\right)\left(n+2\right)...\left(2n\right).\left(2.4.6...2n\right)}\)
\(=\frac{1.2.3...\left(2n-1\right).2n}{1.2.3...2n.2^n}=\frac{1}{2^n}\)
\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right).\left(3n+2\right)}=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right).\left(3n+2\right)}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\left(\frac{3n+2}{2.\left(3n+2\right)}-\frac{2}{2.\left(3n+4\right)}\right)\)
\(=\frac{1}{3}.\frac{3n}{2.\left(3n+2\right)}=\frac{n}{2.\left(3n+2\right)}\)
Ta có \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\) (đpcm)
Áp dụng công thức trên ta có
A\(=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdot\cdot\cdot\cdot\cdot\cdot\cdot+\frac{1}{2015\cdot2016\cdot2017}\)
\(\Leftrightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{2015\cdot2016\cdot2017}\)
\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{2}{3\cdot4}+....+\frac{1}{2015\cdot2016}-\frac{1}{2016\cdot2017}\)
\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2016\cdot2017}\)
\(\Rightarrow A=\left(\frac{1}{1\cdot2}-\frac{1}{2016\cdot2017}\right)\div2\approx0.25\)
Vậy A\(\approx0.25\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)
\(=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)