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\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)...\left(\frac{n^2-1}{n^2}\right)\)
\(=\text{[}\frac{\left(2-1\right)\left(2+1\right)}{2^2}\text{]}.\text{[}\frac{\left(3-1\right)\left(3+1\right)}{3^2}\text{]}.\text{[}\frac{\left(4-1\right)\left(4+1\right)}{4^2}\text{]}...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)
\(=\left(\frac{1.3}{2^2}\right).\left(\frac{2.4}{3^2}\right).\left(\frac{3.5}{4^2}\right)...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)
\(=\frac{\text{[}1.2.3...\left(n-1\right)\text{]}.\text{[}3.4.5...\left(n+1\right)\text{]}}{\text{[}2.3.4...n\text{]}.\text{[}2.3.4...n\text{]}}\)
\(=\frac{1}{n}.\frac{n+1}{2}\)
\(=\frac{n+1}{2n}\)
Với mọi k thuộc N và k > 2 thì ta có :
\(1-\frac{1}{1+2+....+k}=1-\frac{1}{\frac{k\left(k+1\right)}{2}}=1-\frac{2}{k\left(k+1\right)}=\frac{k^2+k-2}{k\left(k+1\right)}=\frac{\left(k+2\right)\left(k-1\right)}{k\left(k+1\right)}\)
Áp dụng vào A ta được :
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+....+n}\right)\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(=\frac{\left[1.2.3....\left(n-1\right)\right]\left[4.5.6.....\left(n+2\right)\right]}{\left(2.3.4......n\right)\left[3.4.5.....\left(n+1\right)\right]}\)
\(=\frac{n+2}{n.3}=\frac{n+2}{3n}\)
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)
\(\left(x^n+1\right)\left(x^n-2\right)-x^{n-3}\left(x^{n+3}-x^3\right)+2018=x^{2n}+x^n-2.x^n-2-x^{2n}+x^n+2018=2016.\)
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 = \(\frac{3.\left(n+1\right)}{n+2}\)