hình như phân số cuối  phải là 1/324

nếu là 1/324 thì tớ giải nè:

A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324

= 1/4.(1+1/2^2+1/3^2+1/4^2+1/5^2+1/6^2+1/7^2+1/8^2+1/9^2)                                                    <1/4.(1+1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9)

= 1/4.(1+1-1/9)

= 1/4.17/9 = 17/36<18/36 = 1/2

=> A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324<1/2

CHO ĐÚNG NHA!!!!!!!!!!!!!!!!

cho dung na

nha bai tren sai day yhemh moi dung ne

A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324

= 1/4.(1+1/2^2+1/3^2+1/4^2+1/5^2+1/6^2+1/7^2+1/8^2+1/9^2) <1/4.(1+1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9)

= 1/4.(1+1-1/9)

= 1/4.17/9 = 17/36<18/36 = 1/2

=> A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324<1/2

1/4+1/16=1\20

1\20+1/36=1/56

1/56+1/64=1\120

1/120+1/100=1/220

1/220+1/44=1/264

1/264+1/196=1/460

1/460+1/256=1/716

suy ra:1/716 ko phải số TN

Ta có : \(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{256}>0\)

\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}+\frac{1}{256}\)

\(A< \frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+\frac{1}{49}+\frac{1}{64}\)\(+\frac{1}{81}+\frac{1}{100}+\frac{1}{121}+\frac{1}{144}+\frac{1}{169}+\frac{1}{196}+\frac{1}{225}+\frac{1}{256}\)

\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{15\cdot16}\)

\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{15}-\frac{1}{16}\)

\(A< 1-\frac{1}{16}< 1\)

\(\Rightarrow0< A< 1\)

=> A ko là số tự nhiên

\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{324}+\frac{1}{400}\)

\(A=\frac{1}{4}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\right)\)

\(< \frac{1}{4}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)=\frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{19}{40}< \frac{20}{40}=\frac{1}{2}\)

đề là < 1/2 nhé

dpcm là gì vậy các bồ

dpcm là điều phải chứng minh nha

Ta có :     \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)

                 \(\frac{1}{16}< \frac{1}{4}-\frac{4}{8}\)

                \(\frac{1}{36}< \frac{1}{8}-\frac{1}{12}\)

                \(\frac{1}{64}< \frac{1}{12}-\frac{1}{16}\)

                 \(\frac{1}{100}< \frac{1}{16}-\frac{1}{20}\)

                   \(\frac{1}{144}< \frac{1}{20}-\frac{1}{24}\)

                 \(\frac{1}{196}< \frac{1}{24}-\frac{1}{28}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{28}\)

             \(=\frac{1}{2}-\frac{1}{28}< \frac{1}{2}\)

Vậy A<1/12

hình như phân số cuối  phải là 1/324

nếu là 1/324 thì tớ giải nè:

A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324

= 1/4.(1+1/2^2+1/3^2+1/4^2+1/5^2+1/6^2+1/7^2+1/8^2+1/9^2)                                                    <1/4.(1+1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9)

= 1/4.(1+1-1/9)

= 1/4.17/9 = 17/36<18/36 = 1/2

=> A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324<1/2

Cảm ơn nha Nobita Kun!!!

A<1/1*2+1/3*4+........+1/17*18

A<1-1/2+1/3-1/4+.......+1/17-1/18

A<(1+1/3+.....+1/17)-(1/2+1/4+......+1/18)

A<(1+1/2+1/3+......+1/18)-(1/2+1/4+.....+1/18)-(1/2+1/4+.......+1/18)

A<1-1/18-(1/2+1/3+1/4+......+1/17)

A<17/18-1/2-(1/3+1/4+......+1/17)

A<4/9-(1/3+1/4+.......+1/17)<1/2=4/8

Vậy a<1/2(đpcm)--------------------------Mình làm hơi dài nhé----------------------------------