N=\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+... +\(\frac{1}{a\left(a+1\right)}\)
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DB
3
7 tháng 10 2019
\(A=-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{\left(n-1\right)n}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n-1.n}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(=-\left(1-\frac{1}{n}\right)\)
\(=-\frac{n-1}{n}\)
S
16 tháng 9 2019
\(A=-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{\left(n-1\right).n}\)
\(A=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\right)\)
\(A=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{\left(n-1\right)}-\frac{1}{n}\right)\)
\(\Rightarrow A=-\left(1-\frac{1}{n}\right)\)
B
3
19 tháng 7 2017
Ta có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{a\left(a+1\right)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{a}-\frac{1}{a+1}\)
\(=1-\frac{1}{a+1}\)
\(=\frac{a+1}{a+1}-\frac{1}{a+1}=\frac{a}{a+1}\)
2 tháng 8 2017
Ta có công thức :
\(\frac{1}{k\left(k+1\right)}=\frac{\left(k+1\right)-k}{k\left(k+1\right)}=\frac{k+1}{k\left(k+1\right)}-\frac{k}{k\left(k+1\right)}=\frac{1}{k}-\frac{1}{k+1}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}=\frac{n-1}{n}\)
NN
27 tháng 3 2017
Ta có:
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{2n+1}{n^2}-\frac{2n+1}{\left(n+1\right)^2}\)
\(=1-\frac{2n+1}{\left(n+1\right)^2}\)
Vậy \(A=\frac{2n+1}{\left(n+1\right)^2}\)
11 tháng 5 2020
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}=\frac{5}{6}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(B=1-\frac{1}{n+1}=\frac{n}{n+1}\)
23 tháng 8 2019
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
23 tháng 8 2019
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
14 tháng 7 2018
Ta có : 1/ 1.2 + 1/ 2.3 + 1/ 3.4 + ... + 1/ n.( n + 1 ) .
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..... + 1/n - 1/ n+1 .
= 1 - 1/ n + 1 .
= n+1 / n+1 - 1/ n+1 .
= n/ n+1 .
Đáp sô : n/ n+1
5 tháng 5 2019
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{5\cdot6}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}\)
\(A=\frac{5}{6}\)
5 tháng 5 2019
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}\)
\(A=\frac{5}{6}\)
\(B=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(B=\frac{100}{2}\)
Thay n = a nha / lúc trước có giải r nên ko giải lại rắc rối
\(N=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{a\left(a+1\right)}\)
\(\Rightarrow N=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{a}-\frac{1}{a-1}\)
\(\Rightarrow N=1-\frac{1}{a-1}\)
\(\Rightarrow N=\frac{a-1-1}{a-1}\)
\(\Rightarrow N=\frac{a-2}{a-1}\)