\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{18}+\frac{1}{30}+\frac{1}{45}+...+\frac{1}{14850}\)

\(\Rightarrow\frac{3}{2}S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

               \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{30}+\frac{1}{45}+...+\frac{1}{14850}\)

\(\Rightarrow\frac{3}{2}S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

               \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

               \(=1-\frac{1}{100}=\frac{99}{100}\)

Vậy S = \(\frac{99}{100}:\frac{3}{2}\) = \(\frac{33}{50}\)

\(F=\left(\frac{3}{1.8}+\frac{3}{8.15}+\frac{3}{15.22}+...+\frac{3}{106.113}\right)\)\(-\)\(\left(\frac{25}{50.55}+\frac{25}{55.60}+\frac{25}{60.65}+...+\frac{25}{95.100}\right)\)

\(=\frac{3}{7}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+...+\frac{1}{106}-\frac{1}{113}\right)\) -  \(5\left(\frac{1}{50}-\frac{1}{55}+\frac{1}{55}-\frac{1}{60}+...+\frac{1}{95}-\frac{1}{100}\right)\)

\(=\frac{3}{7}\left(\frac{1}{3}-\frac{1}{113}\right)-5\left(\frac{1}{50}-\frac{1}{100}\right)\)

\(=\frac{3}{7}.\frac{110}{339}-5.\frac{1}{100}\)

\(=\frac{1}{7}-\frac{1}{20}=\frac{13}{140}\)

= \(\frac{3}{7}\left(\frac{7}{1.8}+\frac{7}{8.15}+...+\frac{7}{106.103}\right)-5\left(\frac{5}{50.55}+\frac{5}{55.60}+...+\frac{5}{95.100}\right)\)

=\(\frac{3}{7}\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+...+\frac{1}{106}-\frac{1}{113}\right)-5\left(\frac{1}{50}-\frac{1}{55}+\frac{1}{55}-\frac{1}{60}+...+\frac{1}{95}-\frac{1}{100}\right)\)

=\(\frac{3}{7}\left(1-\frac{1}{113}\right)-5\left(\frac{1}{50}-\frac{1}{100}\right)\)

=\(\frac{3}{7}\cdot\frac{112}{113}-5\cdot\frac{1}{100}\)

=\(\frac{48}{113}-\frac{1}{20}\)

=\(\frac{847}{2260}\)

So sánh s với 2 biếts=1+13 +16 +110 +............+145 

Như vậy ta sẽ so sánh 1 và 1/3 + 1/6 + 1/10 +......+ 1/45

Ta có :  1/3 + 1/6 + 1/10 + .....+  1/45 <  1/10 + 1/10 + 1/10 +......+  1/10 

Mà 1/10 + 1/10 + 1/10 + ....+  1/10 = 8/10 < 1

     Vậy S <2

S = 1 + 1/3 + 1/6 + 1/10 + ... + 1/45

S = 2/2 + 2/6 + 2/12 + 2/20 + ... + 2/90

S = 2 × (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/9×10)

S = 2 × (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10)

S = 2 × (1 - 1/10) < 2 × 1 = 2

=> S < 2

ta có 1/3=10/30

1/21+1/22+...+1/30 có 10 p/số

mà 1/21>1/30

1/22>1/30

....

1/29>1/30

1/30=1/30

=>1/21+..1/30>1/30+....1/30 có 10 phân số 

=>1/21+...1/30>1/3

Ta có: \(\frac{1}{21}< \frac{1}{30}\)

\(\frac{1}{22}< \frac{1}{30}\)

......

\(\frac{1}{29}< \frac{1}{30}\)

\(\Rightarrow S< \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\)(có 10 p/s)
\(\Rightarrow S< \frac{1}{30}.10=\frac{10}{30}=\frac{1}{3}\)

Vậy S < 1/3

ta co 1/21+1/22+1/23>3/30

1/24+1/25+1/26>3/30

1/27+1/28+1/29>3/30

==>S>3/30+3/30+3/30+1/30

S>10/30 hay S>1/3

\(S=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}+...+\frac{1}{\sqrt{2025}-\sqrt{2024}}\)

Ta nhận xét thấy mỗi số hạng trong S đều dương. Từ đó ta đặt

\(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}+...+\frac{1}{\sqrt{2024}-\sqrt{2023}}\left(A>0\right)\)

\(\Rightarrow S=A+\frac{1}{\sqrt{2025}-\sqrt{2024}}=A+\frac{\sqrt{2025}+\sqrt{2024}}{\left(\sqrt{2025}-\sqrt{2024}\right)\left(\sqrt{2025}+\sqrt{2024}\right)}\)

\(=A+\sqrt{2025}+\sqrt{2024}>\sqrt{2025}=45\)

Vậy \(S>45\)

PS: Phan Thanh Tịnh xem lại bài giải nhé bạn

Ta có : 1 = (n + 1) - n =\(\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2\)

\(=\left(\sqrt{n+1}\right)^2-\sqrt{n+1}.\sqrt{n}+\sqrt{n+1}.\sqrt{n}+\left(\sqrt{n}\right)^2\)

\(=\sqrt{n+1}.\left(\sqrt{n+1}-\sqrt{n}\right)+\sqrt{n}.\left(\sqrt{n+1}-\sqrt{n}\right)\)

\(=\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n-1}+\sqrt{n}\right)\)\

\(\Rightarrow\frac{1}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}+\sqrt{n}\)

Áp dụng vào bài toán,ta có :

\(S=\sqrt{1}+\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2025}-\sqrt{2024}=\sqrt{2025}\)= 45

Vậy S = 45

vgfdgfd

Ta có 

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

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

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

=> S < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

S < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(S< 1-\dfrac{1}{100}< 1\)(do 1/100 >0)

ĐPcm

Giải:

\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\) 

Ta có:

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\) 

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\) 

\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\) 

\(...\) 

\(\dfrac{1}{99^2}=\dfrac{1}{99.99}< \dfrac{1}{98.99}\) 

\(\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\) 

\(\Rightarrow S< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\) 

\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\) 

\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{100}< 1\) 

\(\Rightarrow S< 1\) 

Vậy S < 1.

S=1/5+1/5²+1/5³+...+1/5⁸

5S=1+1/5+1/5²+...+1/5⁷

4S=1-1/5⁸<1

\(\Rightarrow\)S<1

gium nha

S=1/5+1/5²+1/5³+...+1/5⁸

5S=1+1/5+1/5²+...+1/5⁷

4S=1-1/5⁸<1

=> S<1

S=1/5+1/5²+1/5³+...+1/5⁸

5S=1+1/5+1/5²+...+1/5⁷

4S=1-1/5⁸<1

=> S<1

Đáp số: đpcm