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
Những câu hỏi liên quan
Tính tổng A=\(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot5\cdot6}+...+\frac{1}{27\cdot28\cdot29\cdot30}\)
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)
\(A=\frac{1}{4.6}+\frac{1}{10.12}+\frac{1}{18.20}+...+\frac{1}{810.812}\)
.......
~ Chúc học tốt ~
Ai ngang qua xin để lại 1 L - I - K - E
Đúng 0
Bình luận (0)
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{27.28.29.30}\)
\(3A=3.\left(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+......+\frac{1}{27.28.29.30}\right)\)
\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+..........+\frac{3}{27.28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{6}-\frac{1}{24360}\)
\(3A=\frac{1353}{8120}\)
\(A=\frac{1353}{8120}:3\)
\(A=\frac{451}{8120}\)
Đúng 0
Bình luận (0)
Ta có:3A=\(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+.............+\frac{3}{27.28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)
\(3A=\frac{1353}{8120}\Rightarrow A=\frac{451}{8120}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
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
Đúng 0
Bình luận (0)
Cho S_1-S_2+S_3-S_4+S_5frac{m}{n} với m, n nguyên tố cùng nhau. Biết:S_1frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}S_2frac{1}{2cdot3}+frac{1}{2cdot4}+frac{1}{2cdot5}+frac{1}{2cdot6}+frac{1}{3cdot4}+frac{1}{3cdot5}+frac{1}{3cdot6}+frac{1}{4cdot5}+frac{1}{4cdot6}+frac{1}{5cdot6}S_3frac{1}{2cdot3cdot4}+frac{1}{2cdot3cdot5}+frac{1}{2cdot3cdot6}+frac{1}{2cdot4cdot5}+frac{1}{2cdot4cdot6}+frac{1}{2cdot5cdot6}+frac{1}{3cdot4cdot5}+frac{1}{3cdot4cdot6}+frac{1}{3cdot5cdot6}+frac{1}{4cdot5cdot6}...
Đọc tiếp
Cho \(S_1-S_2+S_3-S_4+S_5=\frac{m}{n}\) với m, n nguyên tố cùng nhau. Biết:
\(S_1=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
\(S_2=\frac{1}{2\cdot3}+\frac{1}{2\cdot4}+\frac{1}{2\cdot5}+\frac{1}{2\cdot6}+\frac{1}{3\cdot4}+\frac{1}{3\cdot5}+\frac{1}{3\cdot6}+\frac{1}{4\cdot5}+\frac{1}{4\cdot6}+\frac{1}{5\cdot6}\)
\(S_3=\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot5}+\frac{1}{2\cdot3\cdot6}+\frac{1}{2\cdot4\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{2\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot6}+\frac{1}{3\cdot5\cdot6}+\frac{1}{4\cdot5\cdot6}\)
\(S_4=\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{2\cdot3\cdot4\cdot6}+\frac{1}{2\cdot3\cdot5\cdot6}+\frac{1}{2\cdot4\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5\cdot6}\)
\(S_5=\frac{1}{2\cdot3\cdot4\cdot5\cdot6}\)
Tính \(m+n\)
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
Đúng 0
Bình luận (0)
Tính \(A=\frac{1}{1\cdot2\cdot3\cdot4\cdot5}+\frac{1}{2\cdot3\cdot4\cdot5\cdot6}+...+\frac{1}{26\cdot27\cdot28\cdot29\cdot30}\)
chứng tỏ rằng : \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}<1\)Tính tổng sau : S = \(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{48\cdot49}+\frac{1}{49\cdot50}\)
Xem chi tiết
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}
Đúng 0
Bình luận (0)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
Giúp mk với mk cảm ơn ạ
=1-1/2+1/2-1/3+1/3-1/4+1*4-1/5+1/5+1/6=1-1/6=5/6
ok xong
Đúng 0
Bình luận (0)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}=1-\frac{1}{2}+\)+.... \(+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}=\frac{5}{6}\)
Đúng 0
Bình luận (0)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}\)
\(=\frac{6}{6}-\frac{1}{6}=\frac{5}{6}\)
Ai tk mk mk tk lai nha
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
TÍNH TỔNG:
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+.....+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
Tính hợp lí:
A=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\)
Ta có :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
Vậy \(A=\frac{49}{50}\)
Chúc bạn học tốt ~
Đúng 0
Bình luận (0)
A= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
A= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
A= \(1-\frac{1}{50}\)
A= \(\frac{49}{50}\)
Đúng 0
Bình luận (0)
A= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
A= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
A= \(1-\frac{1}{50}\)
A= \(\frac{50}{50}-\frac{1}{50}\)
A= \(\frac{49}{50}\)
Vậy A= \(\frac{49}{50}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời