Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+......+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
Vậy..
\(B=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{n^2+3n+2-2}{4\left(n+1\right)\left(n+2\right)}=\dfrac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)
\(L_1=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2015.2016.2017}\)
\(L_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}-\dfrac{1}{2016.2017}\right)\)
\(L_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2016.2017}\right)\)
\(L_1=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2016.2017}\right)\)
\(L_1=\dfrac{1}{4}-\dfrac{1}{2.2016.2017}\)
\(L_2=1.2+2.3+...+2006.2007\)
\(3L_2=1.2.3+2.3.\left(4-1\right)+...+2006.2007.\left(2008-2005\right)\)
\(3L_2=1.2.3+2.3.4-1.2.3+...+2006.2007.2008-2005.2006.2007\)\(3L_2=2006.2007.2008\)
\(L_2=\dfrac{2006.2007.2008}{3}\)
\(pt\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{2.2016.2017}\right).x=\dfrac{2006.2007.2008}{3}\)
Dễ dàng tìm được x nhé
Ta có: \(\dfrac{1}{n.\left(n+1\right).\left(n+2\right)}=\dfrac{1}{2}.\left(\dfrac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\right)=\dfrac{1}{2}\left(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(\Rightarrow B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{n^2+3n}{4\left(n+1\right)\left(n+2\right)}\)
Bài 1a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2018.2019}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
b) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)
\(2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2017.2019}\)
\(2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)
\(2S=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
\(S=\dfrac{1009}{2019}\)
Còn lại bạn làm tương tự hết nhé .
Đặt: \(\left\{{}\begin{matrix}l_1=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2005.2006.2007}\\l_2=1.2+2.3+3.4+...+2006.2007\end{matrix}\right.\Leftrightarrow l_1.x=l_2\)
Ta có:
\(l_1=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2005.2006.2007}\)
\(l_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2005.2006}-\dfrac{1}{2006.2007}\right)\)
\(l_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2006.2007}\right)\)
\(l_2=1.2+2.3+3.4+...+2006.2007\)
\(3l_2=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+2006.2007.\left(2008-2005\right)\)
\(3l_2=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2006.2007.2008-2005.2006.2007\)
\(3l_2=2006.2007.2008\Leftrightarrow l_2=\dfrac{2006.2007.2008}{3}\)
Hay: \(\left[\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2006.2007}\right)\right].x=\dfrac{2006.2007.2008}{3}\)
Tới đây thì bấm máy tính là ra :V
Nhã Doanh, ngonhuminh, nguyen thi vang, Hoàng Anh Thư, Mashiro Shiina, Phạm Nguyễn Tất Đạt, F.C, Trần Thị Hồng Ngát, Mến Vũ, kuroba kaito, @Phùng Khánh Linh, Nguyễn Huy Tú, Lightning Farron, Hung nguyen, ...
b.
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+....+\dfrac{2}{\left(n-1\right).n.\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{4}-\dfrac{1}{2n\left(n+1\right)}\)