\(\frac{1}{2}S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{101}}\)

=> \(\frac{1}{2}S-S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}+\frac{1}{2^{101}}-\frac{1}{2^{100}}-...-\frac{1}{2}-1\)

<=> \(\frac{-1}{2}S=\frac{1}{2^{101}}-1\)

<=> \(S=2-\frac{1}{2^{100}}\)

Ta có : 

S = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\left(1\right)\)

\(\Rightarrow2S=2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\left(2\right)\)

Lấy (2) - (1) ta được :

\(S=2-\frac{1}{2^{100}}=\frac{2^{101}-1}{2^{100}}\)

Đăng từ bài thôi bạn à!

a) Áp dụng công thức: \(\frac{1}{a-1}-\frac{1}{a}=\frac{1}{\left(a-1\right)a}>\frac{1}{a.a}=\frac{1}{a^2}\)

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

\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)

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

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

\(\frac{1}{n^2}< \frac{1}{n-1}-\frac{1}{n}\)

___________________________________________

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}=\frac{1}{n+1}< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\) (đpcm)

Ta có:

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

\(\Rightarrow2S=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2S-S=2-\frac{1}{2^{100}}\)

\(\Rightarrow S=2-\frac{1}{2^{100}}\)

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

\(2.S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(2.S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow S=1-\frac{1}{2^{100}}\)

Ta có

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

Áp dụng vào bài toán

\(S=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+.....+\sqrt{100}-\sqrt{99}\)

\(S=\sqrt{100}-\sqrt{1}=9\)

có cách nào khác ko

\(S=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)

\(\Rightarrow S=\sqrt{2}-1+\sqrt{2}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)

\(\Rightarrow S=\sqrt{10}-1\)

\(\Rightarrow Y=9\)

P/s: Bn cướp nick của người khác thì bn trả lại cho người mất nick đi nhé

Ta có:

\(\frac{1}{5^2}>\frac{1}{5.6}\)

\(\frac{1}{6^2}>\frac{1}{6.7}\)

.......

\(\frac{1}{100^2}>\frac{1}{100.101}\)

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)                                                                                                                                                                               \(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)                                                                                                                             = \(\frac{1}{5}-\frac{1}{101}>\frac{1}{5}-\frac{1}{30}=\frac{1}{6}\)                                                                                  \(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{6}\)                              (1)

   Tương tự ta có:

\(\frac{1}{5^2}< \frac{1}{4.5}\)

\(\frac{1}{6^2}< \frac{1}{5.6}\)

......

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

\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)                                                                                                                                                                      \(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

    \(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4}\)                                (2)

                     Từ (1) và (2)

                  \(\Rightarrow\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4}\)                  (đpcm)

     _Chúc_bạn_học_tốt_

thank you bn chúc bạn hok tot nhé

a)A=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

      =\(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

      =\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

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

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

      =\(\frac{-98}{100}=\frac{-49}{50}\)

a) $A= \backslash (\backslash frac\{1\}\{100\}-\backslash frac\{1\}\{100.99\}-\backslash frac\{1\}\{99.98\}-...-\backslash frac\{1\}\{3.2\}-\backslash frac\{1\}\{2.1\}\backslash )$ 

$A=$$\backslash (\backslash frac\{1\}\{100\}-\backslash left(\backslash frac\{1\}\{1.2\}+\backslash frac\{1\}\{2.3\}+...+\backslash frac\{1\}\{98.99\}+\backslash frac\{1\}\{99.100\}\backslash right)\backslash )$$\mathrm{(\; vận\; dụng quy tắc\; dấu\; ngoặc)}$\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

$\mathrm{\backslash (A=\backslash frac\{1\}\{100\}-\backslash left(1-\backslash frac\{1\}\{100\}\backslash right)\backslash ) = \backslash (\backslash frac\{1\}\{100\}-\backslash frac\{99\}\{100\}\backslash )}$ = \(\frac{-98}{100}\) = \(\frac{-49}{50}\)

$\frac{\mathrm{A=1/}}{100}-\frac{\mathrm{1/}}{100.99}-\frac{\mathrm{1/}}{\mathrm{99.98\; -...}}-\frac{\mathrm{1/}}{3.2}-\frac{\mathrm{1/}}{2.1}$

$A=\frac{\mathrm{1/}}{100}-(\frac{\mathrm{1/}}{1.2}+\frac{\mathrm{1/}}{2.3}+...+\frac{\mathrm{1/}}{98.99}+\frac{\mathrm{1/}}{99.100})$

 (vận dụng quy tắc dấu ngoặc)
$A=\frac{\mathrm{1/}}{100}-(1-\frac{\mathrm{1/}}{2}+\frac{\mathrm{1/}}{2}-\frac{\mathrm{1/}}{3}+...+\frac{\mathrm{1/}}{98}-\frac{\mathrm{1/}}{99}+\frac{\mathrm{1/}}{99}-$

$\frac{\mathrm{1/}}{\mathrm{100\; )}}$

 
$A=\frac{\mathrm{1/}}{100}-(1-\frac{\mathrm{1/}}{\mathrm{100)}}$ ) = $\frac{\mathrm{1/}}{100}-\frac{\mathrm{99/}}{100}=\frac{-\mathrm{98/}}{100}=\frac{-\mathrm{49/}}{50}$

/ là phần nha

S=..... (đề bài)

\(\Rightarrow2S=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)

\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)

\(\Rightarrow2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)

\(\Rightarrow S=1-\frac{1}{2^9}=1-\frac{1}{256}=\frac{255}{256}\)

Ôi, trang wed không tự nhận diện được công thức latex. Mình đăng lại bài giải:

a) Ta có

\(4T=\frac{4}{1+\sqrt{5}}+\frac{4}{\sqrt{5}+\sqrt{9}}+...+\frac{4}{\sqrt{2013}+\sqrt{2017}}\)

\(=\frac{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}{\sqrt{5}+1}+...+\frac{\left(\sqrt{2017}+\sqrt{2013}\right)\left(\sqrt{2017}-\sqrt{2013}\right)}{\sqrt{2017}+\sqrt{2013}}\)

\(=\sqrt{5}-1+\sqrt{9}-\sqrt{5}+\sqrt{13}-\sqrt{9}+...+\sqrt{2017}-\sqrt{2013}\)

\(=\sqrt{2017}-1\)

\(\Rightarrow T=\frac{\sqrt{2017}-1}{4}\)

b) Ta có

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{2-1}{\sqrt{2}\sqrt{1}\left(\sqrt{2}+\sqrt{1}\right)}\)

\(=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{2}\sqrt{1}\left(\sqrt{2}+\sqrt{1}\right)}\)

\(=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}\sqrt{1}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)

Tương tự ta có

\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)

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

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

Suy ra

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

\(=1-\frac{1}{10}=\frac{9}{10}\)

a)\[\begin{array}{l}
4T = \frac{4}{{1 + \sqrt 5 }} + \frac{4}{{\sqrt 5  + \sqrt 9 }} + ... + \frac{4}{{\sqrt {2013}  + \sqrt {2017} }}\\
 = \frac{{(\sqrt 5  + 1)(\sqrt 5  - 1)}}{{1 + \sqrt 5 }} + ... + \frac{{(\sqrt {2017}  + \sqrt {2013} )(\sqrt {2017}  - \sqrt {2013} )}}{{\sqrt {2013}  + \sqrt {2017} }}\\
 = \sqrt 5  - 1 + \sqrt 9  - \sqrt 5  + ... + \sqrt {2017}  - \sqrt {2013} \\
 = 1 + \sqrt 5  - \sqrt 5  + \sqrt 9  - \sqrt 9  + ... + \sqrt {2013}  - \sqrt {2013}  + \sqrt {2017} \\
 = 1 + \sqrt {2017} \\
 \Rightarrow T = \frac{{1 + \sqrt {2017} }}{4}
\end{array}\]

Cảm ơn Tran Tu, kết bn đê