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AH
Akai Haruma
Giáo viên
8 tháng 7 2018

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) \(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{n^2}-1\right)\)

\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.....\frac{1-n^2}{n^2}\)

\(=\frac{(-1)(2^2-1)}{2^2}.\frac{(-1)(3^2-1)}{3^2}....\frac{(-1)(n^2-1)}{n^2}\)

\(=(-1)^{n-1}.\frac{(2^2-1)(3^2-1)...(n^2-1)}{2^2.3^2....n^2}\)

\(=(-1)^{n-1}.\frac{(2-1)(2+1)(3-1)(3+1)...(n-1)(n+1)}{2^2.3^2....n^2}\)

\(=(-1)^{n-1}.\frac{(2-1)(3-1)...(n-1)}{2.3...n}.\frac{(2+1)(3+1)...(n+1)}{2.3...n}\)

\(=(-1)^{n-1}.\frac{1.2.3...(n-1)}{2.3...n}.\frac{3.4...(n+1)}{2.3.4...n}\)

\(=(-1)^{n-1}.\frac{1}{n}.\frac{n+1}{2}=(-1)^{n-1}.\frac{n+1}{2n}\)

21 tháng 5 2018

Hơi khó nhìn, bn thông cảm!!!

a: \(=\left(-\dfrac{5}{7}\right)^{n-n}=\left(-\dfrac{5}{7}\right)^0=1\)

b: \(=\left(-\dfrac{1}{2}\right)^{2n-n}=\left(-\dfrac{1}{2}\right)^n\)

8 tháng 10 2017

a) = 1-1/2+1/2-1/3+...+1/99-1/100 =1 - 1/100 = 99/100

3 tháng 9 2017

1, \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...........\left(1-\dfrac{1}{n+1}\right)\)

\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right)...........\left(\dfrac{n+1}{n+1}-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}..............\dfrac{n}{n+1}\)

\(=\dfrac{1.2.3........n}{2.3.......\left(n+1\right)}\)

\(=\dfrac{1}{n+1}\)

2, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...........+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+............+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}\)

\(=\dfrac{99}{100}\)

C=\(-66\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)

=\(-66.\left(\dfrac{5}{66}\right)+124\left(-37-63\right)=-5+124.\left(-100\right)\)

=-12405

23 tháng 7 2018

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) \(\dfrac{\left(-\dfrac{1}{2}\right)^{2n}}{\left(-\dfrac{1}{2}\right)^n}\)

\(=\dfrac{\left(-\dfrac{1}{2}\right)^n.\left(-\dfrac{1}{2}\right)^n}{\left(-\dfrac{1}{2}\right)^n}\)

\(=\left(-\dfrac{1}{2}\right)^n\)

4 tháng 10 2021

\(N=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{100}\)

\(\Rightarrow2N=2+1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\)

\(\Rightarrow N=2N-N=2+1+\dfrac{1}{2}+...+\left(\dfrac{1}{2}\right)^{99}-1-\dfrac{1}{2}-...-\left(\dfrac{1}{2}\right)^{100}=2-\left(\dfrac{1}{2}\right)^{100}\)

4 tháng 10 2021

\(N=1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\)

\(\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\)

\(\dfrac{1}{2}N-N=\left(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\right)\)

               \(-\left(1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\right)\)

\(-\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}-1\)

\(N=\dfrac{-\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}}{-\dfrac{1}{2}}\)