\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)\(\)
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B1 Nhân biểu tthức trên với 1/2 được
1/6+1/12/+1/20+....+1/110=1/2×3+1/3×4+.....+1/10×11
=1/2-1/3+1/3-1/4+.....+1/10-1/11=1/2-1/11=9/22
B2
B.1/1×2+1/2×3+.....+1/99×100=1-1/2 +1/2-1/3+....+1/99-1/100=1-1/100=99/100
Phần a sai đề phải llà1/2×4+1/4×6+.....+1/8×10 mới làm đc nhé
Đặt :
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(\Leftrightarrow\)\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)
\(\Leftrightarrow\)\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)
\(\Leftrightarrow\)\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
\(\Leftrightarrow\)\(A=1-\frac{1}{2^7}\)
Vậy \(A=1-\frac{1}{2^7}\)
1/2+1/4+1/8+1/16+1/32+1/64=32/64+16/64+8/64+4/64+2/32+1/64=63/64<1
Ta có: \(1\frac{4}{5}+2\frac{5}{7}+3\frac{4}{5}+4\frac{5}{7}\)
\(=\left(1\frac{4}{5}+3\frac{4}{5}\right)+\left(2\frac{5}{7}+4\frac{5}{7}\right)\)
\(=\left(\frac{9}{5}+\frac{19}{5}\right)+\left(\frac{19}{7}+\frac{33}{7}\right)\)
\(=\frac{28}{5}+\frac{52}{7}=13\frac{1}{35}\)
= ( \(1\frac{4}{5}\)+ \(3\frac{4}{5}\)) + ( \(2\frac{5}{7}\)+ \(4\frac{5}{7}\))
= \(4\frac{4}{5}\) + \(6\frac{5}{7}\)
= \(\frac{24}{5}\) + \(\frac{47}{7}\)
= ...... ( tính nốt nhé )
gọi biểu thức là A
A=1/2+1/4+1/8+...+1/2048=1/2+1/2^2+1/2^3+...+1/2^10
=>2A=1+1/2+1/2^2+...+1/2^9
=>A=2A-A(bạn đặt cột dọc ra rồi sẽ thấy:1/2-1/2=0;1/2^2-1/2^2=0;...)Ta được kết quả bằng 1+1/2^10
Đặt A =1/2 + 1/4 + 1/8 + ...+ 1/1024 + 1/2048
A= 1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11
2A= 1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10
2A-A= (1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10) - (1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11)
A= 1+1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10 - 1/2 - 1/2^2 - 1/2^3 - ...- 1/2^10 - 1/2^11
A= 1- 1/2^11
A= 2047/ 2048
\(3B=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(B=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4B=3B+B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
+ Đặt \(M=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(3M=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(\Rightarrow4M=3M+M=3-\frac{1}{3^{99}}\)
\(\Rightarrow M=\frac{3}{4}-\frac{1}{3^{99}\cdot4}\)
\(\Rightarrow4B=M-\frac{100}{3^{100}}=\frac{3}{4}-\frac{1}{3^{99}\cdot4}-\frac{100}{3^{100}}\)
\(\Rightarrow B=\frac{3}{16}-\frac{1}{3^{99}\cdot16}-\frac{100}{3^{100}\cdot4}\) \(\Rightarrow B< \frac{3}{16}\)
a) \(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
\(A=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
\(\Rightarrow3A=2A+A=1-\frac{1}{2^6}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{2^6\cdot3}< \frac{1}{3}\) ( đpcm )
\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{4}{14}-\frac{2}{13}}\times\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{\frac{2}{6}+\frac{2}{14}-\frac{2}{26}}{\frac{4}{6}+\frac{4}{14}-\frac{4}{26}}\times\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{64}-\frac{3}{356}}{\frac{4}{4}-\frac{4}{16}+\frac{4}{64}-\frac{4}{256}}+\frac{5}{8}\)
\(=\frac{2\left(\frac{1}{6}+\frac{1}{14}-\frac{1}{26}\right)}{4\left(\frac{1}{6}+\frac{1}{14}-\frac{1}{26}\right)}\times\frac{3\left(\frac{1}{4}-\frac{1}{16}+\frac{1}{64}-\frac{1}{356}\right)}{4\left(\frac{1}{4}-\frac{1}{16}+\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)
\(=\frac{2}{4}\times\frac{3}{4}+\frac{5}{8}\)
\(=\frac{1}{2}\times\frac{3}{4}+\frac{5}{8}\)
\(=\frac{3}{8}+\frac{5}{8}\)
\(=\frac{8}{8}=1\)
\(\frac{\frac{109}{3.7.13}}{\frac{361}{3.14.13}}\)\(\frac{\frac{153}{256}}{\frac{51}{64}}\)+5/8
=\(\frac{327}{722}\)+5/8
=\(\frac{3113}{2888}\)
\(ĐặtA=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
\(A=1-\frac{1}{64}=\frac{63}{64}\)
= 32/64+16/64+8/64+4/64+2/64+1/64
=63/64