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\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)\(<1\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}<1\)
Vậy \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}<1\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}<1\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{47}-\frac{1}{48}+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+....+\frac{1}{25}\right)\)\(=\frac{1}{26}+...+\frac{1}{50}\)
mình chỉ biết câu b thôi:
Ta biến đổi vế phải :
1-1/2+1/3-1/4+.....+1/49-1/50
=(1+1/3+1/5+....+1/49)-(1/2+1/4+1/6+.......+1/50)
=(1+1/2+1/3+.....+1/49+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
Vậy 1/26+1/27+1/28+.....+1/50=1-1/2+1/3-1/4+......+1/49-1/50
A=1 - 1/2 + 1/3 - 1/4 +..+ 1/49 - 1/50
A= 1-( 1/2 + 1/3 ) - ( 1/4 + 1/5 ) -.....-(1/48 + 1/49) - 1/50
A=1 - 5/6 - 9/20 -.....-97/2352 - /150
A= 1 -............cho con lai tu lam nha
A=1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + ... +1/49*50
= 1 - 1/2 + 1/2- 1/3 + ... + 1/49 - 1/50
= 1 - 1/50
= 49/50 <1
A=1 - 1/2+1/2 - 1/3+...+1/49 - 1/50=1/2 - 1/50
=25/50 - 1/50=24/50<1