Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
=>\(2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
=>\(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
=>\(S=1-\frac{1}{2^9}=\frac{511}{512}\)
Vậy \(S=\frac{511}{512}\)
Ta có : \(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^9}\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^8}\)
\(\Rightarrow2S-S=1-\frac{1}{2^9}\)
\(\Leftrightarrow S=1-\frac{1}{2^9}\)
\(S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(2S=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2S-S=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)+\left(1+\frac{1}{2}+...+\frac{1}{2^{10}}\right)\)
\(2S-S=S=2-\frac{1}{2^{10}}\)
\(S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(2S=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(2S=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
\(S=2S-S\)
\(S=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(S=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\)
\(S=2-\frac{1}{2^{10}}\)
Bài 1
a) \(P=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
b) \(S=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{33}{99}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)\(Q=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Tk mình nha!!
Câu 2:
\(P=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right).\left(\frac{3}{3}+\frac{1}{3}\right).\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{99}{99}+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{3\cdot4\cdot5...100}{2.3.4...99}\)
\(=\frac{3\cdot100}{2}\)
\(=\frac{300}{2}=150\)
\(A=\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)
\(\Rightarrow3A=1+\frac{1}{3^0}+\frac{1}{3^1}+...+\frac{1}{3^{2004}}\)
\(\Rightarrow2A=1-\frac{1}{3^{2005}}\)
\(\Rightarrow A=\frac{3^{2005}-1}{3^{2005}.2}\)
a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
... . . . .
\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)
b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Suy ra \(\frac{2}{5}< S\) (1)
Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Từ đó suy ra S < 8/9
Từ (1) và (2) suy ra đpcm
\(S=\dfrac{1}{2^0}+\dfrac{1}{2^1}+...+\dfrac{1}{2^n}\)
=>\(S=1+\dfrac{1}{2}+...+\dfrac{1}{2^n}\)
=>\(2S=2+1+...+\dfrac{1}{2^{n-1}}\)
=>\(2S-S=2+1+...+\dfrac{1}{2^{n-1}}-1-\dfrac{1}{2}-...-\dfrac{1}{2^n}\)
=>\(S=2-\dfrac{1}{2^n}=\dfrac{2^{n+1}-1}{2^n}\)