\(S=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)

Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)

\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(A< \dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\)

\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A< 1-\dfrac{1}{50}\Rightarrow A< 1\)

Ta có \(S=\dfrac{1}{2^2}\left(1+A\right)\)

Ta có

\(A< 1\Rightarrow1+A< 2\Rightarrow S< \dfrac{1}{2^2}.2=\dfrac{1}{2}\)

2/ = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) +......+\(\dfrac{1}{100.101}\)

= 1-\(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)+.........+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)

=1-\(\dfrac{1}{101}\)=...........

mk làm vậy thôi nha

thông cảm

=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{4.5}\)=\(1-\dfrac{1}{2}+....+\dfrac{1}{4}-\dfrac{1}{5}\)

=1-\(\dfrac{1}{5}=\dfrac{4}{5}\)

tương tự

cho minh xin yeu cau de bai

trả hiểu yêu cầu đề bài là j cả

a)Ta có:\(\dfrac{1}{b}-\dfrac{1}{b+1}=\dfrac{b+1-b}{b\left(b+1\right)}=\dfrac{1}{b^2+b}< \dfrac{1}{b^2}\)(do b>1)

\(\dfrac{1}{b-1}-\dfrac{1}{b}=\dfrac{b-b+1}{\left(b-1\right)b}=\dfrac{1}{b^2-b}>\dfrac{1}{b^2}\)(do b>1)

b)Áp dụng từ câu a

=>\(\dfrac{1}{2}-\dfrac{1}{3}< \dfrac{1}{2^2}< \dfrac{1}{1}-\dfrac{1}{2}\)

\(\dfrac{1}{3}-\dfrac{1}{4}< \dfrac{1}{3^2}< \dfrac{1}{2}-\dfrac{1}{3}\)

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

\(\dfrac{1}{9}-\dfrac{1}{10}< \dfrac{1}{9^2}< \dfrac{1}{8}-\dfrac{1}{9}\)

=>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}< S< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

=>\(\dfrac{1}{2}-\dfrac{1}{10}< S< 1-\dfrac{1}{9}\)

=>\(\dfrac{2}{5}< S< \dfrac{8}{9}\)(đpcm)

giúp mk với ;-;"

1/4^2 + 1/5^2 +... + 1/100^2 < 1/3.4 + 1/4.5 +...+ 1/99.100

A=1/3 - 1/4 + 1/4 - 1/5 +...+ 1/99 - 1/100

=1/3 - 1/100 < 1/3

a) =

b) >

c) =

d) =

a) =

b) >

c) =

d) <

a) =

b) >

c) =

d) =

b\()\)

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4

Tương tự như vậy với câu a\()\)

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 1/2

Bài 2 :

\(S=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+............+\dfrac{2017}{4^{2017}}\)

\(\Leftrightarrow4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...........+\dfrac{2017}{4^{2016}}\)

\(\Leftrightarrow4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+..........+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+..........+\dfrac{2017}{4^{2017}}\right)\)

\(\Leftrightarrow3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+.........+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)

Đặt :

\(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2016}}\)

\(\Leftrightarrow4A=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2015}}\)

\(\Leftrightarrow4A-A=\left(4+1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2016}}\right)\)

\(\Leftrightarrow3A=4-\dfrac{1}{4^{2016}}\)

\(\Leftrightarrow D=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}\)

\(\Leftrightarrow3S=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}-\dfrac{2017}{4^{2016}}\)

\(\Leftrightarrow3S< \dfrac{4}{3}\)

\(\Leftrightarrow S< \dfrac{4}{9}\)

\(\Leftrightarrow S< \dfrac{1}{2}\rightarrowđpcm\)

\(A=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\) ( A cho đẹp :v)

\(4A=4\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)

\(4A=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\)

\(4A-A=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)\(3A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)

Đặt:

\(M=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\)

\(4M=4\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)

\(4M=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\)

\(4M-M=\left(4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)\(3M=4-\dfrac{1}{4^{2016}}\)

\(M=\dfrac{4}{3}-\dfrac{1}{4^{2016}}\)

Thay M vào A ta có:

\(A=\dfrac{4}{9}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2017}}\)

\(\Rightarrow A< \dfrac{1}{2}\Rightarrowđpcm\)