chứng minh 1/1x2+1/3x4+...+1/49x50=1/26+1/27+1/28+...+/50
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QV
1
3 tháng 8 2016
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\left(đpcm\right)\)
N
24 tháng 8 2017
\(\frac{1}{1x2}+\frac{1}{3x4}+....+\frac{1}{49x50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)+\left(-\frac{1}{2}-\frac{1}{4}-.....-\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}......+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+......+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)\)
7 tháng 4 2016
1/1x2+1/2x3+...+1/49x50
=1-1/2+1/2-1/3+.....+1/49-1/50
=1-1/50(1)
Ta co 1(2)
So sanh (1) voi (2) ta thay 1-1/50<1
=>1/1x2+...+1/49x50<1
(Phuong phap khu)
7 tháng 4 2016
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
=\(\frac{1}{1}-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}<1\)
Vậy \(\frac{49}{50}<1\)
PT
5
27 tháng 4 2017
Ta có công thức \(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)
Dựa vào công thức trên, ta có
\(\frac{1}{1.2}=\frac{1}{2-1}.\left(1-\frac{1}{2}\right)\)
\(\frac{1}{2.3}=\frac{1}{3-2}.\left(\frac{1}{2}-\frac{1}{3}\right)\)
............................................
\(\frac{1}{49.50}=\frac{1}{50-49}.\left(\frac{1}{49}-\frac{1}{50}\right)\)
\(A=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow A=1-\frac{1}{50}=\frac{49}{50}\)
chắc chắn bạn ạ, ai thấy đúng hì ủng hộ nha
27 tháng 4 2017
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{50}=\frac{49}{50}\)\(\frac{49}{50}\)
PH
7
15 tháng 4 2018
Ta có :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
Vậy \(A=\frac{49}{50}\)
Chúc bạn học tốt ~
NH
15 tháng 4 2018
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50
= 1/1 - 1/50
= 49/50
HY
1
LT
2
6 tháng 4 2016
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
6 tháng 4 2016
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{50}\)
\(\frac{50}{50}-\frac{1}{50}=\frac{49}{50}\)
Vì \(\frac{49}{50}<1\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}<1\)
NN
6
28 tháng 4 2016
khó quá sai thì thôi nha
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/49 - 1/50
= 1/1 - 1/50
= 49/50
28 tháng 4 2016
1/1.2 + 1/2.3 + 1/3.4 +.....+ 1/49.50
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +......+ 1/49 - 1/50
= 1 - 1/50
= 49/50
DT
5
\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
= \(\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)\)
= \(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
Vậy \(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\) = \(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\) hay
\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49}-\dfrac{1}{50}\) = \(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)