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\(B=\frac{5}{1.2.3}+\frac{5}{2.3.4}+...+\frac{5}{n.\left(n+1\right)\left(n+2\right)}\)
\(\Leftrightarrow\frac{2B}{5}=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow B=\frac{5}{4}-\frac{5}{2\left(n+1\right)\left(n+2\right)}\)
A = \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+........+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
A = \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+............+\frac{2n+1}{2^2.\left(n+1\right)^2}\)
A = \(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+........\frac{2n+1}{n^2.\left(n+1\right)^2}\)
A = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+.........+\frac{2n+1}{n^2}-\frac{2n+1}{\left(n+1\right)2}\)
A = \(\frac{1}{1}-\frac{2n+1}{\left(n+1\right)^2}\)
A = \(1-\frac{2n+1}{\left(n+1\right)2}\)
nha bạn.
B=1/2.1.2-1/2.2.3+1/2.2.3-1/2.3.4+...+1/2n(n+1)-1/2(n+1)(n+2)
B=1/2[(1/1.2+1/2.3+...+1/n(n+1))-(1/2.3+1/3.4+...+1/(n+1)(n+2))]
Tới đây bạn tự làm tiếp nha, tương tự như bài 1/1.2+1/2.3+..+1/n(n+1) á bạn.Cái này bạn ghi ra bạn sẽ hiểu, mình viết hơi bị lủng củng.
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{9.10}\right)\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)
\(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{9}-\frac{1}{10}\right)\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)
\(\Rightarrow\left(1-\frac{1}{10}\right)\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)
\(\Rightarrow\frac{9}{10}.\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)
\(\Rightarrow\frac{9}{10}x-\frac{9}{10}+\frac{1}{10}x=x-\frac{9}{10}\)
\(\Rightarrow\left(\frac{9}{10}x+\frac{1}{10}x\right)-\frac{9}{10}=x-\frac{9}{10}\)
\(\Rightarrow x-\frac{9}{10}=x-\frac{9}{10}\)
\(\Rightarrow x\inℝ\)
Vậy \(x\inℝ\)
Em chỉ làm những bài e biết thôi, thông cảm nhs :D
a/ chịu
b/ \(C=1+7+7^2+.........+7^{50}\)
\(\Leftrightarrow7C=7+7^2+...........+7^{50}+7^{51}\)
\(\Leftrightarrow7C-C=\left(7+7^2+.......+7^{51}\right)-\left(1+7+.....+7^{50}\right)\)
\(\Leftrightarrow6C=7^{51}-1\)
\(\Leftrightarrow C=\dfrac{7^{51}-1}{6}\)
c/ \(A=\dfrac{-1}{4}+\dfrac{7}{3}+\dfrac{3}{4}+\dfrac{9}{2}\)
\(=\left(\dfrac{-1}{4}+\dfrac{3}{4}\right)+\left(\dfrac{7}{3}+\dfrac{9}{2}\right)\)
\(=\dfrac{1}{4}+\dfrac{41}{6}\)
\(=\dfrac{85}{12}\)
d/ Thấy phép tính hơi dài
e/ \(C=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+.........+\dfrac{1}{2015.2016.2017}\)
\(\Leftrightarrow2C=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+.........+\dfrac{2}{2015.2016.2017}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.......+\dfrac{1}{2015.2016}-\dfrac{1}{2016.2017}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2016.2017}\)
\(=\dfrac{1}{2}-\dfrac{1}{4066272}\)
\(=\dfrac{2033136}{4066272}\)
\(\Leftrightarrow C=\dfrac{2033136}{4066272}:2\)
\(\Leftrightarrow C=?\)
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)...\left(\frac{n^2-1}{n^2}\right)\)
\(=\text{[}\frac{\left(2-1\right)\left(2+1\right)}{2^2}\text{]}.\text{[}\frac{\left(3-1\right)\left(3+1\right)}{3^2}\text{]}.\text{[}\frac{\left(4-1\right)\left(4+1\right)}{4^2}\text{]}...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)
\(=\left(\frac{1.3}{2^2}\right).\left(\frac{2.4}{3^2}\right).\left(\frac{3.5}{4^2}\right)...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)
\(=\frac{\text{[}1.2.3...\left(n-1\right)\text{]}.\text{[}3.4.5...\left(n+1\right)\text{]}}{\text{[}2.3.4...n\text{]}.\text{[}2.3.4...n\text{]}}\)
\(=\frac{1}{n}.\frac{n+1}{2}\)
\(=\frac{n+1}{2n}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}\)
kb nha
ok