những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé

A=\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+\(\frac{1}{5^2}\)+...+\(\frac{1}{98^2}\)

A=\(\frac{1}{3.3}\)+\(\frac{1}{4.4}\)+\(\frac{1}{5.5}\)+...+\(\frac{1}{98.98}\)

A<\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{97.98}\)=\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+...+\(\frac{1}{97}\)-\(\frac{1}{98}\)=\(\frac{1}{2}\)-\(\frac{1}{98}\)=\(\frac{24}{49}\)<1.

Vậy A<1

Ta có: 

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

                                                                       \(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

                                                                       \(=\frac{1}{1}-\frac{1}{100}\)

                                                                        \(=\frac{99}{100}\)

Mà \(\frac{99}{100}< 1\)                                                                        

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)

Vậy \(A< 1\)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

\(\frac{1}{4^2}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)

...

\(\frac{1}{100^2}=\frac{1}{100\cdot100}< \frac{1}{99\cdot100}\)

=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

=> \(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

=> \(A< \frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

Lại có : \(\frac{99}{100}< 1\)

=> \(A< \frac{99}{100}< 1\)=> \(A< 1\)( đpcm )

Đặt A=1/10+1/40+1/88+1/154+1/238+1/340

A=1/2.5+1/5.8+1/8.11+1/11.14+1/14.17+1/17.20

3A=3/2.5+3/5.8+....+3/17.20

3A=1/2-1/5+1/5-1/8+...+1/17-1/20

3A=1/2-1/20

3A=9/20

2)

Giữ nguyên p/s 1/2^2

Ta có:1/3^2<1/2.3

         1/4^2<1/3.4

        ...............

          1/n^2<1/(n-1).n

=>1/3^2+1/4^2+...+1/n^2<1/2.3+1/3.4+...+1/(n-1).n

=>1/3^2+1/4^2+.....+1/n^2<1/2-1/3+1/3-1/4+.........+1/n-1-1/n

=>1/2^2+1/3^2+.....+1/n^2<1/2^2+1/2-1/n

=>1/2^2+1/3^2+....+1/n^2<3/4-1/n<3/4

3)

2B=2/3.5+2/5.7+....+2/47.49+2/49.51

2B=1/3-1/5+1/5-1/7+.....+1/47-1/49+1/49-1/51

2B=1/3-1/51

2B=16/51

B=16/51:2

B=8/51

A=1+1/2+1/2^2+...+1/2^2010

2A=2+1+1/2+....+1/2^2009

2A-A=(2+1+1/2+...+1/2^2009)-(1+1/2+1/2^2+....+1/2^2010)

A=2-1/2^2010

Ta có:

\(A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2^2A=1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)

\(\Rightarrow4A=1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)

\(\Rightarrow4A-A=1-\frac{1}{2^{100}}< 1\Rightarrow3A< 1\Rightarrow A< \frac{1}{3}\left(đpcm\right)\)

A=221​+241​+261​+...+21001​

\(\Rightarrow 2^{2} A = 1 + \frac{1}{2^{2}} + \frac{1}{2^{4}} + . . . + \frac{1}{2^{98}}\)

\(\Rightarrow 4 A = 1 + \frac{1}{2^{2}} + \frac{1}{2^{4}} + . . . + \frac{1}{2^{98}}\)

\(\Rightarrow 4 A - A = 1 - \frac{1}{2^{100}} < 1 \Rightarrow 3 A < 1 \Rightarrow A < \frac{1}{3} \left(\right. đ p c m \left.\right)\)