CMR: \(\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{97}{48^2\cdot49^2}+\frac{99}{49^2\cdot50^2}< 1\)
Bài này mình chắc 100%, 1 đúng nha vì ghi cực khổ lắm:
1) Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}...+\frac{50-49}{49.50}\)
\(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+...+\frac{50}{49.50}-\frac{49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}
tính\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...\frac{1}{48\cdot49\cdot50}\)
Gọi tổng trên là A
A = 1/1.2.3 + 1/2.3.4 +......+ 1/48.49.50
2A = 2/1.2.3 + 2/2.3.4 +.......+ 2/48.49.50
2A = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 +.......+ 1/48.49 - 1/49.50
2A = 1/1.2 - 1/49.50
2A = 1224/2450
A = 612/2450 = 306/1225
A =\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+.........+\frac{1}{48\cdot49\cdot50}\)
Vậy A =........
Ths các pạn trc na!!!!!
ta thấy 1/(1*2)-1/(2*3)=1/3=2*1/(1*2*3)
do đó A=1/2*{[1/(1*2)-1/(2*3)+[1/(2*3)-1/(3*4)]+.....+[1/(48*49)-1/(49*50)]}
=1/2*[1/(1*2)-1/(2*3)+1/(2*3)-1/(3*4)+.....+1/(48*49)-1/(49*50)]
=1/2*[1/(1*2)-1/(49*50)]
=1/2*(1/2-1/2450)
=1/2*612/1225
=306/1225
Tính: \(B=\frac{100^2+1^2}{100\cdot1}+\frac{99^2+2^2}{99\cdot2}+\frac{98^2+3^2}{98\cdot3}+...+\frac{52^2+49^2}{52\cdot49}+\frac{51^2+50^2}{51\cdot50}\)
Tính Tổng :
\(A=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot5\cdot6}+...+\frac{1}{47\cdot48\cdot49\cdot50}\) mọi người giúp em với ạ
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+....+\frac{1}{47.48.49.50}\)
\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{47.48.49}-\frac{1}{48.49.50}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{48.49.50}\right)\)
\(=\frac{1}{3}.\frac{6533}{39200}=\frac{6533}{117600}\)
Tính tổng :\(D=\frac{1}{1^2\cdot2^2}+\frac{1}{2^2\cdot3^2}+\frac{1}{3^2\cdot4^2}+....+\frac{1}{49^2\cdot50^2}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
tinh
Tách: 1/1.2=1-1/2; 1/2.3=1/2-1/3; ....; 1/49.50=1/49-1/50
Và rút gọn các số liền kề thì còn lại kết quả
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(=\left(1-\frac{1^1}{2}\right)+\left(\frac{1^1}{2}-\frac{1^1}{3}\right)+\left(\frac{1^1}{3}-\frac{1}{4}^1\right)+...+\left(\frac{1^1}{49}-\frac{1}{50}\right)\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
Cho \(A=1\cdot3\cdot5\cdot7\cdot...\cdot49\)
\(B=\frac{1\cdot2\cdot3\cdot...\cdot48\cdot49\cdot50}{2\cdot4\cdot6\cdot...\cdot48\cdot50}\)
\(C=\frac{26}{2}\cdot\frac{27}{2}\cdot...\cdot\frac{50}{2}\)
So sánh A,B và C
TRẢ LỜI ĐI CÓ DC KO ĐỂ MK CÒN GIẢI
A=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{49\cdot50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
A=1/1.2+1/2.3+...+1/49.50
=1-1/2+1/2-1/3+...+1/49-1/50
=1-1/50
=49/50
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(\Rightarrow A=\frac{49}{50}\)