Cho A = 1/1.2 + 1/3.4 + 1/5.6 +...+ 1/49.50
B = 1/1 + 1/2 + 1/3 + 1/4 + ... + 1/49 + 1/50
C = 1/2 + 1/4 + 1/6 +...+1/48 + 1/50
CMR : A = B - 2C
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- Ta có B-2C = (1/1+1/2+1/3+1/4+...+1/49+1/50) - (1/1+1/2+1/3+....+1/25)
= 1/26+1/27+1/28+...+1/50
- Ta có A= 1-1/1.2+1/3-1/4+1/5-1/6+.....+1/49-1/50
= (1+1/2+1/3+14+...+1/49)-(1/2+1/4+1/6+...+1/50)
=(1+1/2+1/3+...+1/50)-2(1/2+1/4+1/6+...+1/50)
= ( 1+1/2+...+1/50)-(1+1/2+1/3+...+1/25)
= 1/26+1/27+....+1/50
=> A=B-2C <ĐPCM>
Vậy A=B-2C
- mik giải hơi dài theo cách cổ điển ^_^ ^_^
A = 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/49.50
A =1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/49 - 1/50
A = (1 + 1/3 + 1/5 + ... + 1/49) - (1/2 + 1/4 + 1/6 + ... + 1/50)
A = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/49 + 1/50) - 2.(1/2 + 1/4 + 1/6 + ... +1/50)
A = B - 2C
=> đpcm
Ủng hộ mk nha ♡_♡☆_☆
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{50}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(A=B-2C\left(đpcm\right)\)
A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
A = \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
A = \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
A = B - 2C ( ĐPCM )
Vậy A = B - 2C