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\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{49.50}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}\)
\(M=\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+........+\left(-\frac{1}{49}+\frac{1}{49}\right)-\frac{1}{50}\)
\(M=\frac{1}{1}-0+0+0+0+0+......+0+0-\frac{1}{50}\)
\(M=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
Vì \(\frac{49}{50}<1\) nên \(S<1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
M=1/1.2+1/2.3+...+1/49.50
M=1/1-1/2+1/2-1/3+.....+1/49-1/50
M=1-1/50<1
=>M<1
\(M=\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{49.50}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(M=1-\frac{1}{50}<1\)
\(=>M<1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(M=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{49}+\frac{1}{49}\right)-\frac{1}{50}\)
\(M=1+0+0+...+0-\frac{1}{50}\)
\(M=\frac{49}{50}\)
\(\Rightarrow\frac{49}{50}< 1\)
\(\Rightarrow M< 1\)
dấu chấm ở giữa hai số là dấu nhân à?
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
Vì \(\frac{49}{50}< 1\)\(\Rightarrow M< 1\)
VẬY M < 1
HK TỐT #
\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}< 1\)
\(\Leftrightarrow M< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
M\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{49}-\frac{1}{50}\)
\(1-\frac{1}{50}=\frac{49}{50}\)
vì \(\frac{49}{50}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(B=1.2+2.3+3.4+...+49.50\)
\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(=49.50.51\)
\(B=\frac{49.50.51}{3}=49.50.17\)
\(50^2.A-\frac{B}{17}=49.50-49.50=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
thấy công thức trên vào biểu thức, khử liên tiếp, ta con
1-1/50 <1
Ta cộng vào biểu thức trên( đặt là A) 1 dãy là:1/2*3+1/4*5+1/6*7+...+1/47*48.(đặt là B).
=>A+B>A.
Ta có:A+B= 1/1*2+1/2*3+1/3*4+1/4*5+...+1/49*50.
=>A+B=1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50.
=>A+B=1-1/50.
=>A+B<.
Mà A+B>A=>A<1.
Vậy A<1.
tk nha đúng 1000000% .
-chúc các bạn tk mk học giỏi nha-
![](https://rs.olm.vn/images/avt/0.png?1311)
hình như ko phải so sánh mà là còn cái nịt (:
M =1/1.2+1/2.3+....+1/49.50
M=1/1-1/2+1/2-1/3+...+1/49-1/50
M=1/1-1/50
M=49/50
tính nha :-)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt A=1/1.2+1/2.3+1/3.4+..........1/49.50
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}<1\)
vậy A<1
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 < 1
= 1/1-1/2+1/2-1/3+...+1/49-1/50
= 1/1-1/50
= 50/50-1/50
= 49/50
Mà 1 = 50/50
Nên 49/50 < 50/50
Vậy M < 1
m=1/1.2+1/1.3 +...+1/49 .50
=1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50
=1-1/50 <1 (vi1/50 >0)