?

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

Bạn muốn tính toán giá trị của E hay muốn so sánh E với một số khác?

Đặt A=\(\frac{1}{3}.5+\frac{1}{5}.7+...+\frac{1}{97}.99\)

=>A=\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

=>2A=\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

=>2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

=>2A=\(\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)

=>A=\(\frac{32}{99}:2=\frac{32}{99}.\frac{1}{2}=\frac{32}{198}=\frac{16}{99}\)

\(A=\frac{1}{2}-\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^4+...-\left(\frac{1}{2}\right)^{20}\)

\(2A=1-\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3+...-\left(\frac{1}{2}\right)^{19}\)

\(2A-A=\)\(\left(1-\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3+...-\left(\frac{1}{2}\right)^{19}\right)-\)\(\left(\frac{1}{2}-\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^4+...-\left(\frac{1}{2}\right)^{20}\right)\)

\(A=1-\left(\frac{1}{2}\right)^{20}\)

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

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^_^

\(\frac{1}{2}.2^n+2^{2+n}=9.2^5\)

\(\frac{1}{2}.2^n+4.2^n=9.2^5\)

\(2^n.\left(\frac{1}{2}+4\right)=9.2^5\)

\(2^n.\frac{9}{2}=9.2^5\)

\(2^n=9.\frac{2}{9}.2^5\)

\(2^n=2.2^5\)

\(2^n=2^6\)

\(\Rightarrow n=6\)

\(\frac{1}{2}\)\(\times\)\(2^n\)\(+\)\(2^{2+n}\)\(=\)\(9\)\(\times\)\(2^5\)

\(\frac{1}{2}\)\(\times\)\(2^n\)\(\times\)\((\)\(1\)\(+\)\(2^2\)\()\)\(=\)\(9\)\(\times\)\(2^5\)

\(\frac{1}{2}\)\(\times\)\(2^n\)\(\times\)\(5\)\(=\)\(9\)\(\times\)\(2^5\)

\(2^n\)\(=\)\(9\)\(\times\)\(32\)\(\div\)\(5\)\(\times\)\(2\)

\(2^n\)\(=\)115,2

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

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

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

=> A < B