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5 tháng 4 2020

Đặt A = \(\left(1+\frac{2}{4}\right).\left(1+\frac{2}{10}\right).\left(1+\frac{2}{18}\right).....\left(1+\frac{2}{n^2+3n}\right)\)

Ta có : A = \(\left(1+\frac{2}{4}\right).\left(1+\frac{2}{10}\right).\left(1+\frac{2}{18}\right).....\left(1+\frac{2}{n^2+3n}\right)\)

                 \(\frac{6}{4}.\frac{12}{10}.\frac{20}{18}.....\frac{\left(n+1\right).\left(n+2\right)}{n.\left(n+3\right)}\)

                = \(\frac{3.2}{4}.\frac{3.4}{2.5}.\frac{4.5}{3.6}.....\frac{\left(n+1\right).\left(n+2\right)}{n.\left(n+3\right)}\)

                = \(\frac{3.2.3.4.4.5....n}{2.3.4.5.6.....\left(n+2\right)}\)

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

Vậy A = \(\frac{3.\left(n+1\right)}{n+2}\)

22 tháng 11 2016

a) \(\frac{\left(n+1\right)!}{n!\left(n+2\right)}=\frac{n!\left(n+1\right)}{n!\left(n+2\right)}=\frac{n+1}{n+2}\)

b)\(\frac{n!}{\left(n+1\right)!-n!}=\frac{n!}{n!\left(n+1\right)-n!}=\frac{n!}{n!\left(n+1-1\right)}=\frac{1}{n}\)

c)\(\frac{\left(n+1\right)!-\left(n+2\right)!}{\left(n+1\right)!+\left(n+2\right)!}=\frac{n!\left(n+1\right)-n!\left(n+1\right)\left(n+2\right)}{n!\left(n+1\right)+n!\left(n+1\right)\left(n+2\right)}=\frac{n!\left(n+1\right)\left(1-n-2\right)}{n!\left(n+1\right)\left(1+n+2\right)}=\frac{-n-1}{n+3}\)

( Kí hiệu n!=1.2.3.4...n)

22 tháng 11 2016

cảm ơn bạn nhiều nhiều nhiều lắm

14 tháng 12 2018

\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)

  \(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)...\left(\frac{n^2-1}{n^2}\right)\)

\(=\text{[}\frac{\left(2-1\right)\left(2+1\right)}{2^2}\text{]}.\text{[}\frac{\left(3-1\right)\left(3+1\right)}{3^2}\text{]}.\text{[}\frac{\left(4-1\right)\left(4+1\right)}{4^2}\text{]}...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)

\(=\left(\frac{1.3}{2^2}\right).\left(\frac{2.4}{3^2}\right).\left(\frac{3.5}{4^2}\right)...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)

\(=\frac{\text{[}1.2.3...\left(n-1\right)\text{]}.\text{[}3.4.5...\left(n+1\right)\text{]}}{\text{[}2.3.4...n\text{]}.\text{[}2.3.4...n\text{]}}\)

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

\(=\frac{n+1}{2n}\)

15 tháng 1 2017

3/4x8/9x15/16x24/25x....x(n^2-1)/n^2)

=(1x2x3x4x...x(n-1))x(3x4x5x...x(n+1)):(1x2x3x4x...x n)^2

=(n+1)/2n)

15 tháng 1 2017

Ta có công thức :

\(1-\frac{1}{k^2}=\frac{k^2-1^2}{k^2}=\frac{\left(k+1\right)\left(k-1\right)}{k^2}\)

Áp dụng công thức trên ta được :

\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)

\(=\frac{2^2-1^2}{2^2}.\frac{3^2-1^2}{3^2}.\frac{4^2-1^2}{4^2}....\frac{n^2-1^2}{n^2}\)

\(=\frac{\left(2+1\right)\left(2-1\right)}{2.2}.\frac{\left(3+1\right)\left(3-1\right)}{3.3}.\frac{\left(4+1\right)\left(4-1\right)}{4.4}...\frac{\left(n+1\right)\left(n-1\right)}{n.n}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{\left(n+1\right)\left(n-1\right)}{n.n}\)

\(=\frac{\left[1.2.3.....\left(n+1\right)\right].\left[3.4.5...\left(n-1\right)\right]}{\left(2.3.4....n\right)\left(2.3.4....n\right)}\)

\(=\left(n+1\right).\frac{1}{2n}=\frac{n+1}{2n}\)

15 tháng 1 2017

mới lớp 6 mà giải đc toan lớp 8 , anh đây thuông minh quá ))