Bài 2

a) Ta có

S = \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

S = \(\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Vì \(\dfrac{1}{13}< \dfrac{1}{12}\)

\(\dfrac{1}{14}< \dfrac{1}{12}\)

\(\dfrac{1}{15}< \dfrac{1}{12}\)

=> \(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}.3\)

Lại có

\(\dfrac{1}{61}< \dfrac{1}{60}\)

\(\dfrac{1}{62}< \dfrac{1}{60}\)

\(\dfrac{1}{63}< \dfrac{1}{60}\)

=> \(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}.3\)

=> S = \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\) < \(\dfrac{1}{5}+\dfrac{1}{12}.3+\dfrac{1}{60}.3\)

= \(\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\) = \(\dfrac{1}{2}\)

=> đpcm

Ta có

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{2015}{2016}\)

\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{2015}{2016}\)

\(\dfrac{1}{1}-\dfrac{1}{x+2}=\dfrac{2015}{2016}\)

\(\dfrac{1}{x+2}=\dfrac{1}{1}-\dfrac{2015}{2016}\)

\(\dfrac{1}{x+2}=\dfrac{1}{2016}\)

2016 = x + 2

x = 2016 - 2

x = 2014

Vậy x = 2014 là giá trị cần tìm

b) Ta có

S = 1/2 + 1/2² + 1/2³ + ... + 1/2²⁰

2S = 1 + 1/2 + 1/2² + ... + 1/2¹⁹

=> 2S - S = 1 - 1/2²⁰

S = 1 - 1/2²⁰ < 1

=> Ta có đpcm

              \(1/1.3+1/3.5+1/5.7+...+1/n.(n+2)<2003/2004\)

Ta có :=2/2.(1/1.3+1/3.5+1/5.7+...+1/n.(n+2)

           =1/2.(2/1.3+2/3.5+2/5.7+...+2/n.(n+2)

           =1/2.(1-1/3+1/3-1/5+1/5-1/7+...+1/n-1/n+2)

           =1/2.(1-1/n+2)

           =1/2.(n+2/n+2-1/n+2)

           =1/2.(n+2-1/n+2)

           =1/2.n+1/n+2

           =n+1/(n+2).2

       Vì: n+1/(n+2).2<2003/2004

Suy ra:n+1/(n+2).2=x/2004

Suy ra:(n+2).2=2004

            n+2     =1002

            n         =1000

Vậy n bằng 1000

S = 1/11 + 1/12 + 1/13 + 1/14 + ... + 1/20

S > 1/20 + 1/20 + 1/20 + 1/20 + ... + 1/20

               10 phân số 1/20

S > 10 × 1/20

S > 1/2

ta thấy: 1/11;1/12;1/13;...;1/19;1/20 đều >1/20

=>1/11+1/12+...1/19+1/20>1/20+1/20...+1/20

1/11+1/12+...1/19+1/20>10/20

1/11+1/12+...1/19+1/20>1/2 vậy S>1/2

vì 1/11>1/20

     1/12>1/20...

     1/13>1/20

nên 1/11+1/12+,,,,+1/20>1/20+1/20+,,,,+1/20=10/20=1/2(rút gọn

                                    10 số 1/20

vậy S>1/2

\(\frac{1}{11}\)> \(\frac{1}{20}\)

\(\frac{1}{12}\)> \(\frac{1}{20}\)

.

.

.

\(\frac{1}{19}\)>\(\frac{1}{20}\)

\(\frac{1}{20}\)= \(\frac{1}{20}\)

=> S  = 1/11+1/12+...+1/20>1/20+1/20+1/20+1/20+1/20+1/20+1/20+1/20+1/20+1/20=10*1/20=1/2 (đpcm)

Ta có:

\(\frac{1}{11}>\frac{1}{20}\)

\(\frac{1}{12}>\frac{1}{20}\)

.............

\(\frac{1}{20}=\frac{1}{20}\) 

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\) ( 10 phân số \(\frac{1}{20}\))

\(\Leftrightarrow\frac{10.1}{20}=\frac{10}{20}=\frac{1}{2}\)

Vì \(\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{2}\). Mà \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{2}\)

Số lượng số của S là : 

\(\left(20-11\right):1+1=10\)( số ) 

Ta có : 
\(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};...;\frac{1}{19}>\frac{1}{20};\frac{1}{20}=\frac{1}{20}\) 

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)

\(\Rightarrow S>\frac{1}{20}.10\)

\(\Rightarrow S>\frac{1}{2}\)

Vậy \(S>\frac{1}{2}\)

Ta có:

 1/11 + 1/12 + 1/13 + ....................... + 1/ 20 > 1/20 +1/20 +1/ 20 +1/20 +1/20 +1/20 +1/20 +1/ 20 +1/20 +1/20 = 1/2 

=> S > 1/2 

             Vậy S > 1/2

\(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};............;\frac{1}{20}=\frac{1}{20}\)    nên cái tổng ấy sẽ > 10 lần 1/20 =1/2

Ta có S = 1/11+1/12+1/13+...+1/19+1/20 nên S có 10 số hạng 
Và 1/2 = 10/20 = 
Mà 1/11 > 1/12 > 1/13 > 1/14 > 1/15 > 1/16 > 1/17 > 1/18 > 1/19 > 1/20 
Nên 1/11+1/12+1/13+...+1/19+1/20 > 1/20x10 
=> 1/11+1/12+1/13+...+1/19+1/20 > 10/20 
=> 1/11+1/12+1/13+...+1/19+1/20 > 1/2 
Vậy S > 1/2

lồn

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