Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(Cm:\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
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:
\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
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 \(\frac{1}{5}< K< \frac{1}{3}.\)
chứng minh rằng 1 phần 2 mũ 2 cộng 1 phần 3 mũ 2 + 1 4 mũ 2 chấm chấm chấm 1 phần 100 mũ 2 nhỏ hơn 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
1/1002 + 1/1012 + ... + 1/1992 < 1/99.100 + 1/100.101 + ... + 1/198.199 = 1/99 - 1/100 + 1/100 - 1/101 + ... + 1/198 - 1/199 = 1/99 - 1/199
\(\Rightarrow\)Vậy 1/1002 + 1/1012 + ... + 1/1992 < 1/99 (vì 1/99 đã lớn hơn 1/99 - 1/199 rồi mà G lại còn bé hơn 1/99 - 1/199 nữa)
1/1002 + 1/1012 + ... + 1/1992 > 1/100.101 + ... + 1/199.200 = 1/100 - 1/101 + ... + 1/199 - 1/200 = 1/100 - 1/200 = 1/200
\(\Rightarrow\)Vậy 1/1002 + 1/1012 + ... + 1/1992 > 1/200
A =\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{20^2}=\frac{1}{2^2}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\right)\)
\(< \frac{1}{2^2}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\right)=\frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(=\frac{1}{4}\left(1+1-\frac{1}{20}\right)=\frac{1}{4}\left(2-\frac{1}{20}\right)=\frac{1}{2}-\frac{1}{80}< \frac{1}{2}\left(\text{đpcm}\right)\)
\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
Gọi A = \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
=> A = \(\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
A < \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
A < \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
A < \(\frac{1}{2}-\frac{1}{100}\)
A < \(\frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)
=> A < \(\frac{1}{2}\)
<=> \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}\)