\(Cm:\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)

Gọi biểu thức trên là A, ta có:

3A = 1-2/3+3/3^2-...-100/3^99

3A + A = [1-2/3+3/3^2-...-100/3^99] + [1/3-2/3^2+3/3^3-...-100/3^100]

4A = 1 - 1/3 + 1/3^2 - ... - 1/3^99 - 100/3^99 [1]

Gọi B = 1-1/3 + 1/3^2 - ... - 1/3^99

3B = 3 - 1 + 1/3 - 1/3^2 -...-1/3^2012

3B + B = [3-1+1/3-1/3^2-...-1/3^2012] + [1-1/3 + 1/3^2 - ... - 1/3^99]

4B = 3 - 1/3^99 

=> 4B < 3 => B < 1/4 [2]

Từ [1], [2] => 4A < B < 3/4 => A < 3/16 [đpcm]

MỎI TAY QUỚ

tk nha

Lúc đặt câu hỏi, bạn bấm vào góc trên cùng bên trái để gõ phép tính đẹp. Ý của bạn có phải là:

\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)

Bạn kia làm dúng rồi^_^

3 nhân 2/3 bao nhiêu

Ta có : \(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)

           \(\frac{1}{5^2}=\frac{1}{5.5}< \frac{1}{4.5}\)     

           \(\frac{1}{6^2}=\frac{1}{6.6}< \frac{1}{5.6}\)

            ...

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

\(\Rightarrow\)K<\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

K<\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

K<\(\frac{1}{3}-\frac{1}{100}< \frac{1}{3}\)

\(\Rightarrow K< \frac{1}{3}\)  (1)

Ta có : \(\frac{1}{4^2}=\frac{1}{4.4}=\frac{1}{16}\)

            \(\frac{1}{5^2}=\frac{1}{5.5}>\frac{1}{5.6}\)

            \(\frac{1}{6^2}=\frac{1}{6.6}>\frac{1}{6.7}\)

             ...

             \(\frac{1}{99^2}=\frac{1}{99.99}>\frac{1}{99.100}\)

             \(\frac{1}{100^2}=\frac{1}{100.100}>\frac{1}{100.101}\)

\(\Rightarrow K>\frac{1}{16}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}+\frac{1}{100.101}\)

K>\(\frac{1}{16}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

K>\(\frac{1}{16}+\frac{1}{5}-\frac{1}{101}>\frac{1}{5}\)  (2)

Từ (1) và (2)

\(\Rightarrow\frac{1}{5}< K< \frac{1}{3}\)

Vậy \(\frac{1}{5}< K< \frac{1}{3}.\)

?

Gỉa sử\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)

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

=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=>\(A< 1-\dfrac{1}{100}\)
=>\(A< \dfrac{99}{100}\)
Mà \(\dfrac{99}{100}< 1\)
=>A<1
Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)

1/2^2 + 1/3^2 + ...+ 1/100^2 
Ta có : 1/2^2 < 1/1.2
1/3^2 < 1/2.3
...
1/100^2 < 1/99.100
=> 1/2^2 + ...+1/100^2 < 1/1.2+1/2.3+...+1/99.100
                                        = 1 - 1/2+1/2-1/3+1/3+...+1/99-1/100
                                        =  1 - 1/100 <1
-> 1/2^2 + ...+1/100^2 < 1

\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-9999}{100^2}\)

\(=-\frac{3.8...9999}{2^2.3^2...100^2}=-\frac{1.3.2.4...99.101}{2.2.3.3...100.100}=-\frac{\left(1.2....99\right).\left(3.4...101\right)}{\left(2.3...100\right).\left(2.3...100\right)}=-\frac{1.101}{100.2}=-\frac{101}{200}\)

\(< -\frac{100}{200}=\frac{1}{2}=B\)

=> A < B