Tìm n biết :\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+......+\(\frac{2}{n.\left(n+1\right)}\)=\(\frac{2003}{2004}\)
Những câu hỏi liên quan
Tìm số tự nhiên n biết: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
Tìm số tự nhiên n biết: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n.\left(n+1\right)}=\frac{2003}{2004}\)
đặt a=1/3+1/6+1/10+...........+2/n(n+1)
1/2a=1/6+1/12+...........+1/n(n+1)
1/2a=1/2.3+1/3.4+........+1/n(n+1)
1/2a=1/2-1/3+1/3-1/4+.......+1/n-1/n+1
1/2a=1/2-1/n+1
a=(1/2--1/n+1):1/2=2003/2004
1/2-1/n+1=2003/2004.1/2
1/2-1/n+1=2003/4008
1/n+1=1/2-2003/4008
1/n+1=1/4008
suy ra n+1=4008
n=4007
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Tìm số tự nhiên n biết :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.........+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
dành cho bạn giỏi toán nâng cao
Tim n để
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{n\left(n+1\right)}=\frac{2003}{2004}\)
Tìm số tự nhiên n biết: \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{10}\)+......+ \(\frac{2}{n.\left(n+1\right)}\)= \(\frac{2003}{2004}\)
Giúp mình với mình đang cần gấp!
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{n\left(n+1\right)}\)
\(=\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{n\left(n+1\right)}=1-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+...+\frac{2}{n}-\frac{2}{n+1}\)
Tới đây dễ rồi bạn rút gọn rồi tìm n
Đúng 0
Bình luận (0)
1 tìm x thuộc N biết
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...................+\frac{1}{x\left(x+1\right):2}=\frac{2001}{2003}\)
Bài 2a) Aleft(frac{1}{2}-1right)left(frac{1}{3}-1right)...left(frac{1}{2002}-1right)left(frac{1}{2003}-1right)b) Bleft(-1frac{1}{2^2}right)left(-1frac{1}{3^2}right)left(-1frac{1}{4^2}right)...left(-1frac{1}{2003^2}right)left(-1frac{1}{2004^2}right)c) Cleft(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(nin N,nge2right)
Đọc tiếp
Bài 2
a) \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)
b) \(B=\left(-1\frac{1}{2^2}\right)\left(-1\frac{1}{3^2}\right)\left(-1\frac{1}{4^2}\right)...\left(-1\frac{1}{2003^2}\right)\left(-1\frac{1}{2004^2}\right)\)
c) \(C=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\left(n\in N,n\ge2\right)\)
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)
\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2001}{2002}\right)\left(-\frac{2002}{2003}\right)\)
\(=\frac{-1.\left(-2\right).....\left(-2001\right)\left(-2002\right)}{2.3....2002.2003}\)
\(=\frac{1}{2003}\)
Đúng 0
Bình luận (0)
Tìm \(x\in N\)biết
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)
\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(-\frac{1}{x+1}=\frac{2001}{4006}-\frac{1}{2}\)
\(-\frac{1}{x+1}=-\frac{1}{2003}\)
\(\Rightarrow x+1=2003\)
\(\Rightarrow x=2012\)
Đúng 0
Bình luận (0)
Ta có: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{2003}:2\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{2003}{4006}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2003}{4006}-\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2}{4006}=\frac{1}{2003}\)
=> x + 1 = 2003
=> x = 2002
Vậy x = 2002
Duyệt nha !!!
chúc hk tốt!!!
Đúng 0
Bình luận (0)
Tìm x biết
a) (8-5x).(x+2)+4.(x-2).(x+1)+2.(x-2).(x+2)=0
b)\(\left(-\frac{2}{5}+x\right):\frac{7}{9}+\left(-\frac{3}{5}+\frac{5}{6}\right):\frac{7}{9}=0\)
c)\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2004}\)