Các câu hỏi tương tự
Tính:
A=(1-1/1+2).(1-1/1+2+3).(1-1/1+2+3+4)...(1-1/1+2+3+4+...+2022)
B=1+1/2(1+2)+1/3(1+2+3)+1/100(1+2+3+...+100)
G=(1-1/2^2).(1-1/3^2).(1-1/4^2).......(1-1/100^2)
H=(1-1/2^2).(1-1/3^2).(1-1/4^2)........(1-1/2018^2)
1^2/1x2x2^2/2x3x4^2/3x5............20^2/19x21 đề bài là tính tích
Bài 2 : Tính tích A = ( 1- 1/1+2)(1-1/1+2+3)(1-1/1+2+3+4).........(1-1/1+2+........+2020)
B = (1/2-1/3)(1/2-1/5)(1/2-1/7).......(1/2-1/99)
tính(1+2/1)(1+2/2)(1+2/3).....(1+2/26)(1+2/27)
(1-1/2^2)(1-1/3^2)(1-1/4^2)........(1-1/2007^2)
1/ cmr:1/2*5+1/5*8+...+1/(3n-1)(3n-2)=n/6n+4
2/ tính:
A=1/2-1/2^2+1/2^3-1/2^4+1/2^5-...+1/2^99-1/2^100
B=(1/2+1)(1/3+1)(1/4+1)...(1/99+1)
B(1-dfrac{1}{2^2})(1-dfrac{1}{3^2})(1-dfrac{1}{4^2})(1-dfrac{1}{5^2})...(1-dfrac{1}{99^2})(1-dfrac{1}{100^2})C(dfrac{1}{4}-1)(dfrac{1}{9}-1)(dfrac{1}{16}-1)...(dfrac{1}{999}-1)
Đọc tiếp
B=(1-\(\dfrac{1}{2^2}\))(1-\(\dfrac{1}{3^2}\))(1-\(\dfrac{1}{4^2}\))(1-\(\dfrac{1}{5^2}\))...(1-\(\dfrac{1}{99^2}\))(1-\(\dfrac{1}{100^2}\))
C=(\(\dfrac{1}{4}\)-1)(\(\dfrac{1}{9}\)-1)(\(\dfrac{1}{16}\)-1)...(\(\dfrac{1}{999}\)-1)
nhận xét dùm coi mk làm đúng không nha :
Rút gọn biểu thức A=1+1/2+1/2^2+1/2^3+...+1/2^2012
=> 2A = 1/2^1+1/2^2+...+1/2^2012+1/2^2013
=> 2A - A = 1A =(1/2^1+1/2^2+...+1/2^2012+1/2^2013)-(1/2^0+1/2^1+1/2^2+1/2^3+...+1/2^2012)
=> 1A = 1/2^2012-1/2^0
= 1/2^2012-1/1
mấy bạn coi ntn
Bài 3:
R=(1-1/3).(1-1/6).(1-1/10)......(1-1/780)
T=(1/2-1).(1/3-1).(1/4-1).....(1/100-1)
K=(1/2^2-1).(1/3^2-1).(1/4^2-1).....(1/100^2-1)
Cm
a) 1/5+1/13+1/14+1/15+1/61+1/62+1/63 nhỏ hơn 1/2
b) 1/2 +1/2^2 +1/2^3 +....+1/2^20 nhỏ hơn 1
c) 1/4+1/5+1/6+...+1/19 lớn hơn 1
d) 3/1.4+3/4.7 +3/7.10 +...+3/40.43+3/43.46 nhỏ hơn 1
e) 1/2^2 + 1/3^2 +1/4^2 +1/5^2 +1/6^2+1/7^2+1/8^2 nhỏ hơn 1
tính:
A=1+!/2+1/2^2+...+1/2^100
B=1+1/3+1/3^2+...=1/3^100
C=(1+1/2)+(1+1/3)=...=(1+1/2018)
D=(1-1/2).(1-1/3)...(1-1/2018)