CM:1-1/2+1/3-1/4+1/5-1/6+.....+!/99-1/100=1/101+1/102+.+1/200
Những câu hỏi liên quan
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.Chưng minh rằng
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
Xét: (1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100) =
(1+1/3+1/5+....+1/99) + (1/2+1/4+1/6+...+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1+1/2+1/3+...+1/50) =
1/51+1/52+1/53+ … + 1/100
Hay:
(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
Viết lại:
(1+1/3+1/5+ … +1/199) – (1/2+1/4+1/6+ … +1/200) = 1/101+1/102+ … +1/200
Tương tự như trên ta được:
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1/2+1/4+1/6+...+1/200) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1+1/2+1/3+...+1/100) =
1/101+1/102+ … +1/200
Hay:
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
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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
Xét: (1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100) =
(1+1/3+1/5+....+1/99) + (1/2+1/4+1/6+...+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1+1/2+1/3+...+1/50) =
1/51+1/52+1/53+ … + 1/100
Hay:
(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
Viết lại:
(1+1/3+1/5+ … +1/199) – (1/2+1/4+1/6+ … +1/200) = 1/101+1/102+ … +1/200
Tương tự như trên ta được:
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1/2+1/4+1/6+...+1/200) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1+1/2+1/3+...+1/100) =
1/101+1/102+ … +1/200
Hay:
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
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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.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.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.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.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.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.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.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
ai tích mình tích lại
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Xem thêm câu trả lời
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
CMR: 1 - 1/2+ 1/3 -1/4+......+1/99-100 = 1/101 + 1/102+......+1/200
Bạn tham khảo lời giải tại đây:
Câu hỏi của Nguyễn Kim Chi - Toán lớp 7 | Học trực tuyến
Và lưu ý lần sau gõ đề bằng công thức toán nhé.
CMR: 1-1/2+1/3-1/4+.....+1/99-1/100 = 1/101 +1/102 +....+ 1/200
Tính các tổng sau:
a) A = 1*2+2*3+3*4+...+2014*2015
b) B = 101^2+102^2+...+199^2+200^2
c) C = 1*3+2*4+3*5+4*6+...+99*101+100*102
cho mi sửa lại:
\(a) A = 1^2+2^3+3^4+...+2014^{2015} b) B = 101^2+102^2+...+199^2+200^2 c) C = 1^3+2^4+3^5+4^6+...+99^{101}+100^{102}\)
dấu 8 là nhân còn dấu ^ là mũ ạ
C/minh:
1-1/2+1/3-1/4+...+1/99-1/100=1/101+1/102+...+1/200
câu 1 : tính tổng
A = 1+(-3) +5+ (-7) + ...+ 17 + (-19 )
C = 1+2-3-4+5+6 -7-8 +...-99-200+101+102
B = 1-4+7-10+...-100+103
A=1+(-3)+5+(-7)+...17+(-19)
=> A=(1+5+9+13+17)-(3+7+11+15+19)
=>A=45-55
=>A=-10
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Ta có :
A=1+(-3)+5+(-7)+...17+(-19)
=> A=(1+5+9+13+17)-(3+7+11+15+19)
=>A=45-55
=>A=-10
Đap số : -10
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CMR : ( 1 + 1/3 + 1/5 + ...+ 1/99 ) + ( 1/2 + 1/4+ ..1/200 ) = 1/101 + 1/102+..+1/200
P/s : Đề sai mik sửa lại rồi : Tham khảo nhé :
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
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\(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{200}-2.\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{200}-1+\frac{1}{2}+....+\frac{1}{100}\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
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bài 1
A=1*2*3+2*3*4+3*4*5+...+99*100*101
B=1*3*5+3*5*7+...+95*97*99
C=2*4+4*6+..+98*100
D=1*2+3*4+5*6+...+99*100
E=1^2+2^2+3^2+...+100^2
G=1*3+2*4+3*5+4*6+...+99*101+100*102
H=1*2^2+2*3^2+3*4^2+...+99*100^2
I=1*2*3+3*4*5+5*6*7+7*8*9+...+98*99*100
K=1^2+3^2+5^2+...+99^2
A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101
=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4
=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)
=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101
=> 4A = 99*100*101*102
=> 4A = 101989800
=> A = 25497450