eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

a) 1 + 3 + 3² + 3³ + ... + 3¹¹

= (1 + 3 + 3² + 3³) + ... + (3⁸ + 3⁹ + 3¹⁰ + 3¹¹)

= 40 + ... + 3⁸.(1 + 3 + 3² + 3³)

= 40 + ... + 3⁸. 40

= (1 + ... + 3⁸) . 40 \(⋮\)40

b) Ta có: B = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) 

        =>    B = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)< \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

      => B <\(1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-...-\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)

     => B < \(1-\frac{1}{100}\)

    => B < 1

\(\frac{1}{1^2}=1\)

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

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

...

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

=> \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1+1-\frac{1}{100}=2-\frac{1}{100}< 2\)

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

Đặt :

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

Ta thấy :

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

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

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

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

\(\Leftrightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...........+\frac{1}{99.100}\)

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

\(\Leftrightarrow A< 1-\frac{1}{100}< 1\)

\(\Leftrightarrow A< 1\)

Ta có 

\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)

\(2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)

Suy ra \(A=2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)

Đặt \(n=\frac{1}{2}\) thì \(A=1+n+n^2+...+n^{99}-\frac{100}{2^{100}}\)

Xét \(B=1+n+n^2+...+n^{99}\Leftrightarrow B.n=n+n^2+n^3+...+n^{100}\)

\(\Leftrightarrow B.n=\left(1+n+n^2+...+n^{99}\right)+\left(n^{100}-1\right)\)

\(\Leftrightarrow B.n=B+n^{100}-1\Leftrightarrow B\left(n-1\right)=n^{100}-1\Leftrightarrow B=\frac{n^{100}-1}{n-1}\)

Suy ra \(A=\frac{\frac{1}{2^{100}}-1}{\frac{1}{2}-1}-\frac{100}{2^{100}}=2\left(1-\frac{1}{2^{100}}\right)-\frac{100}{2^{100}}=-\frac{102}{2^{100}}+2< 2\)

Vậy A < 2

Ta có : 

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

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

Vậy \(P=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)

Chúc bạn học tốt ~

bạn nói với cô giáo là :

bài này nhìn là đủ biết không cần phải  chứng minh

tử số bé hơn mẫu số gần trăm lần :)  éo bao giờ > 1 được :)

Ta có :

\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)

Ta có:

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

Do đó:

\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)

\(\Leftrightarrow A=\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{n-1}+\sqrt{n}\)

\(\Leftrightarrow A=\sqrt{n}-1\left(dpcm\right)\)

Nhan xet:

\(\frac{1}{2^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

\(\frac{1}{3^2}< \frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)

...

\(\frac{1}{100^2}< \frac{1}{100.101}=\frac{1}{100}-\frac{1}{101}\)

Vay: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)

\(P< \frac{1}{2}-\frac{1}{101}=\frac{99}{202}< 1\)

1/2^2 < 1/(1.2)= 1-1/2 
1/3^2 <1/(2.3)=1/2-1/3 
1/4^2 <1/(3.4)=1/3-1/4 
...... 
1/100^2 < 1/99-1/100 
cộng vế với vế ta được 1/2^2 +1/3^2+...< 1-1/2+1/2-1/3+....+1/99-1/100=1-1/100 
=> ĐPCM