Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
5/2 + 5/4 + 5/8 + 5/16 + 5/32 + 5/64
= 5(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64)
Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64
=> 2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32
=> 2A - A = 1 - 1/64
=> A = 1 - 1/64
Do đó : 5(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64) = 5(1 - 1/64) = 5 . 63/64 = 315/64
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
\(1+\frac{5}{4}+\frac{5}{8}+\frac{5}{16}+\frac{5}{32}+\frac{5}{64}\)
\(=\frac{64}{64}+\frac{80}{64}+\frac{40}{64}+\frac{20}{64}+\frac{10}{64}+\frac{5}{64}\)
\(=\frac{64+80+40+20+10+5}{64}\)
\(=\frac{219}{64}\)
\(=\frac{27}{8}\)
1+5/4+5/8+5/16+5/32+5/64
=1+5/4+5/8+5/16+5/32+5/64
=1+(5/4+5/8+5/16+5/32+5/64)
=1+[5x(1/4+1/8+1/16+1/32+1/64)]
A=1/4+1/8+1/16+1/32+1/64
2A=1/2+1/4+1/8+1/16+1/32
2A-A=(1/2+1/4+1/8+1/16+1/32)+(1/4+1/8+1/16+1/32+1/64)
A=1/2-1/64
A=31/64
1+[5x31/64]
=1+155/64
=219/64
TRA LOI:
1+5/4+5/8+5/16+5/32+5/64=219/64
3+3/5+3/25+3/125+3/512=3,8
a) \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)
\(\Rightarrow\)\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)
\(\Rightarrow\)\(2A-A=\left(2+1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^6}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)\)
\(\Leftrightarrow\)\(A=2-\frac{1}{2^7}=\frac{255}{128}\)
b) \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\frac{2}{7}=\frac{1}{7}\)
\(E=\frac{5}{4}+\frac{5}{8}+\frac{5}{16}+\frac{5}{32}+\frac{5}{64}\)
\(\Leftrightarrow E=\frac{5}{2^2}+\frac{5}{2^3}+\frac{5}{2^4}+\frac{5}{2^5}+\frac{5}{2^6}\)
\(\Leftrightarrow E=5\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\right)\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)
\(\Rightarrow2A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\)
\(\Rightarrow2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\right)\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^6}\)
Thay \(A=\frac{1}{2}-\frac{1}{2^7}\)vào E ta được:
\(E=5\cdot\left(\frac{1}{2}-\frac{1}{2^6}\right)\)
Bài làm
~ Đề là tính E, mà làm theo cách của bạn Vũ Hà My đây thì nó lại vừa dài, vừa khó ra kết quả. Nên mik sẽ làm theo cách quy đồng nhé. Dấu " . " là dấu nhân nha. ~
\(E=\frac{5}{4}+\frac{5}{8}+\frac{5}{16}+\frac{5}{32}+\frac{5}{64}\)
\(E=5.\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
\(E=5.\left(\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{32}+\frac{1}{64}\right)\)
\(E=5.\frac{31}{64}\)
\(E=\frac{155}{64}\)
Vậy \(E=\frac{155}{64}\)