Các câu hỏi tương tự
Chứng tỏ rằng :
(1+1/3+1/5+1/7+......+1/101)-(1/2+1/4+1/6+...+1/100) = 1/52+1/53+1/54+.....+1/100+1/101+1/102
CMR: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{101}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+ \frac{1}{102}\right)=\frac{1}{52}+\frac{1}{53}+...+\frac{1}{102}\)
Chứng minh :
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{101}\right)\)\(-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}+\frac{1}{102}\right)\)\(=\)\(\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}+\frac{1}{101}+\frac{1}{102}\)
Mik đng cần gấp , giúp mik nha, giải kĩ cho mik nha
1.Chưng minh rằng (1+/1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh 1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
1. [ ( -2 ). 3 + 9 ]. ( -5 ) - ( -6 ). 8
2. ( 135 - 35 ). ( -47 ) + 53. ( -48 - 52 )
3. 0 - 1 + 2 - 3 + 4 - 5 + 6 - 7 + ... + 2004 - 2005
4. 1 - 3 + 5 - 7 + 9 - 11 + ... + 2005 - 2007
5.1 + 2 + 3 - 4 - 5 - 6 + 7 + 8 + 9 - 10 - 11 - 12 + ... + 97 + 98 + 99 - 100 - 101 - 102
A = 1 . 2 + 2 . 3 + 3 . 4 + ......... + 98 . 99 / 1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + ........... + ( 1 + 2 + 3 + ...... + 98 )
B = ( 1 / 51 . 52 ) + 1 / 52 . 53 + ...... + 1 / 100 . 101 ) : ( 1 / 1 . 2 + 1 / 2 . 3 + ........ + 1 / 99 . 100 + 1 / 100 . 101
1+(-2)+(-3)+4+(-5)+(-6)+...+100+(-101)+(-102)
Xem chi tiết
a) A= 1+(-2)+(-3)+4+5(-6)+(-7)+8+9+...+99+100-101+102+103
b) B=1+(-3)+5+(-7)+...+57+(-99)+101
Tính:
a, A= 1+(-2)+(-3)+4+5+(-6)+(-7)+8+...+99-100-101+102+103
b,B=1+(-3)+5+(-7)+...+97+(-99)+101