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+ \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{102}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{102}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{102}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{102}\right)-\left(1+\frac{1}{2}+...+\frac{1}{51}\right)\)
\(A=\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{102}\)
+ \(154B=\frac{52+102}{52\cdot102}+\frac{53+101}{53\cdot101}+...+\frac{102+52}{102\cdot52}\)
\(154B=\frac{1}{52}+\frac{1}{102}+\frac{1}{53}+\frac{1}{101}+...+\frac{1}{101}+\frac{1}{53}+\frac{1}{102}+\frac{1}{52}\)
\(154B=2\left(\frac{1}{52}+\frac{1}{53}+...+\frac{1}{102}\right)\)
\(B=\frac{1}{77}\left(\frac{1}{52}+\frac{1}{53}+...+\frac{1}{102}\right)\)
Do đó : \(\frac{A}{B}=\frac{1}{\frac{1}{77}}=77\) là số nguyên
A = 1/(1.2) + 1/(3.4) + 1/(5.6) +....+ 1/(1997.1998) =
(1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 1997 - 1 / 1998) =
(1 + 1 / 2 + 1 / 3 + ... + 1998) - 2(1 / 2 + 1 / 4 + ... + 1 / 1998) =
(1 + 1 / 2 + 1 / 3 + ... + 1998) - (1 + 1 / 2 + ... + 1 / 999) =
1 / 1000 + 1 / 1001 + ... + 1 / 1998
2A = (1 / 1000 + 1 / 1001 + ... + 1 / 1998) + (1 / 1998 + 1 / 1997 + ... + 1 / 1000) =
(1 / 1000 + 1 / 1998) + (1 / 1001 + 1 / 1997) + ... + (1 / 1998 + 1 / 1000) =
2998*[1 / (1000*1998) + 1 / (1001*1997) + ... + 1 / (1998*1000)] = 2998B
=> A / B = 1499 nguyên
A = (1/1.2) + (1/3.4) + (1/5.6) +....+ ( 1/1997.1998)
ta có
1/1*2 = 1 - 1/2
1/3*4 = 1/3 - 1/4
...
1/1997*1998 = 1/1007 - 1/1998
bạn gộp lại tự giải tiếp nha
a=\(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{100}=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\frac{1}{51}+...+\frac{1}{100}\)
=>b/a=2011
hình như đề : CMR : \(\frac{b}{a}\)là 1 số nguyên
Ta có :
\(a=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(a=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(a=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(a=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(a=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(a=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(b=\frac{2011}{51}+\frac{2011}{52}+\frac{2011}{53}+...+\frac{2011}{100}\)
\(b=2011.\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)\)
\(\Rightarrow\frac{b}{a}=\frac{2011.\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}=2011\)là 1 số nguyên ( đpcm )
AI NHANH MÌNH K CHO!
Mình mới lớp 4 mà bạn tra mạng sẽ có đầy bài dạng này và y như thế này.