(-2)*(-1+1/2)*(-1+1/3)*(-1+1/4)*....*(-1+1/2001)*(-1+1/2002)
VT
Những câu hỏi liên quan
A= 1/2003×2002 - 1/2002×2001- 1/2001×2000- .....-1/3×2- 1/2×1
Cho A=1^22+1^32+1^42+....+1^20012+1^20022
Chứng tỏ rằng A <2001^2002
S = 1 - 1/2 + 1/3 - 1/4 + ... + 1/2001 - 1/2002 P = 1/1002 + 1/1003 + ... + 1/2002 Hỏi S - P = ?
S=\(\left(1+\frac{1}{2}+......+\frac{1}{2002}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+..........+\frac{1}{2002}\right)\)
=\(\left(1+\frac{1}{2}+.........+\frac{1}{2002}\right)-\left(1+\frac{1}{2}+.........+\frac{1}{1001}\right)\)
=\(\frac{1}{1002}+\frac{1}{1003}+...........+\frac{1}{2002}=P\)
\(\Rightarrow S-P=0\)
Đúng 0
Bình luận (0)
cmr: 1 - 1/2 + 1/3 -1/4 +...-1/200 + 1/2001 - 1/2002 = 1/1002 +...+ 1/2002
AI GIÚP NHÉ!!! THANKS:)
A=2000+2001/2001+2002
B1/2+1/3+1/4+1/5+.....1/99+1/100
cho n! = 1*2*3*...*n ; tinh
A= (1/2! + 2/3! + 3/4! +...+ 2001/2002!) + 1/2002!
\(A=\left(\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{2002-1}{2002!}\right)+\frac{1}{2002!}\)
\(A=\left(\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{2002}{2002!}-\frac{1}{2002!}\right)+\frac{1}{2002!}\)
\(A=\left(\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{2001!}-\frac{1}{2002!}\right)+\frac{1}{2002!}\)
\(A=\frac{1}{1!}-\frac{1}{2002!}+\frac{1}{2002!}=1\)
Đúng 0
Bình luận (0)
Bài 1: 65/303 . -48/102 + 65/102
Bài 2: 1- 1/2 + 1/3 - 1/4 + ... + 1/2001 - 1/2002
Bài 3: 1/1002 + 1/1003 + ..... + 1/2002
Thanks
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2001}-\frac{1}{2002}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2001}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2002}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2002}-1-\frac{1}{2}-...-\frac{1}{1001}\)
\(=\frac{1}{1002}+\frac{1}{1003}+\frac{1}{1004}+...+\frac{1}{2002}\)
Đúng 0
Bình luận (0)
a, chứng minh rằng : 1-1/2+1/3-1/4+...-1/2000+1/2001-1/2002 = 1/1002+ ...+ 1/2002
giúp mk nha
Xem bài tại link này nhé! Bài làm đúng đã đc OLM chọn.
Câu hỏi của Cristiano Ronaldo - Toán lớp 7 - Học toán với OnlineMath
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....-\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+......+\frac{1}{2001}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2001}+\frac{1}{2002}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2002}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{1001}\right)\)
\(=\frac{1}{1002}+\frac{1}{1003}+\frac{1}{1004}+.....+\frac{1}{2002}\)
Chúc em học tốt nhé!
giúp mk bài nữa nha
A=1-1/2+1/3-1/4+...+1/2001-1/2002