1/2-1/2010.2011

cậu có biết tách ko?

nếu cậu biết tách ra thành cách hiệu thì sẽ làm được nhanh thôi

S=1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 +...+ 1/2010.2011 - 1/2011.2012

S=1/1.2 - 1/2011.2012<1/2

=>S<P

75:x=3(du 3 )

\(S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+........+\frac{2}{2010.2011.2012}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+......+\frac{1}{2010.2011}-\frac{1}{2011.2012}\)

\(=\frac{1}{1.2}-\frac{1}{2011.2012}\)

\(=\frac{1}{2}-\frac{1}{2011.2012}\)

mà \(\frac{1}{2}-\frac{1}{2011.2012}< \frac{1}{2}\)

\(\Rightarrow S< P\)

\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)

\(A=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)

A = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{37.38.39}\)

A = \(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{39-37}{37.38.39}\)

A = \(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+....+\frac{39}{37.38.39}-\frac{37}{37.38.39}\)

A = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{37.18}-\frac{1}{38.39}\)

A = \(\frac{1}{2}-\frac{1}{38.39}\)

A = \(\frac{370}{741}\)

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

\(2A=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

mình áp dụng công thức tổng quát:\(\frac{a}{n\left(n+1\right)\left(n+2\right)...\left(n+a\right)}=\frac{1}{n\left(n+1\right)\left(n+a-1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)...\left(n+a\right)}\)

Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)

<=>\(2A=2\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\right)\)

<=>\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

<=>\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

<=>\(2A=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}=\frac{n^2+3n}{2\left(n+1\right)\left(n+2\right)}=\frac{n\left(n+3\right)}{2\left(n+1\right)\left(n+2\right)}\)

<=>\(A=\frac{n\left(n+3\right)}{2\left(n+1\right)\left(n+2\right)}.\frac{1}{2}=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)

tổng quát:  1/n(n+1)(n+2)=1/2[1/n(n+1) - 1/(n+1)(n+2)]

ai k dung mik giai cho