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NN
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
6 tháng 5 2022
Đặt tổng là A
\(2xA=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{64}+\dfrac{1}{128}\)
\(\Rightarrow A=2xA-A=1-\dfrac{1}{256}=\dfrac{255}{256}\)
ND
Nguyễn Đức Trí
VIP
11 tháng 7 2023
a) \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{128}+\dfrac{1}{128}-\dfrac{1}{256}\)
\(=1-\dfrac{1}{256}\)
\(=\dfrac{255}{256}\)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{13.14}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}\)
\(=1-\dfrac{1}{14}\)
\(=\dfrac{13}{14}\)
c) \(\dfrac{3}{15.18}+\dfrac{3}{18.21}+\dfrac{3}{21.24}+...+\dfrac{3}{87.90}\)
\(=3.\left(\dfrac{1}{15.18}+\dfrac{1}{18.21}+\dfrac{1}{21.24}+...+\dfrac{1}{87.90}\right)\)
\(=3.\left[\dfrac{1}{3}.\left(\dfrac{1}{15}-\dfrac{1}{18}\right)+\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{21}\right)+\dfrac{1}{3}.\left(\dfrac{1}{21}-\dfrac{1}{24}\right)+...+\dfrac{1}{3}.\left(\dfrac{1}{87}-\dfrac{1}{90}\right)\right]\)
\(=3.\dfrac{1}{3}.\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{24}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)
\(=\dfrac{1}{15}-\dfrac{1}{90}\)
\(=\dfrac{6}{90}-\dfrac{1}{90}\)
\(=\dfrac{5}{90}=\dfrac{1}{18}\)
29 tháng 4 2015
Đặt A=1/2+1/4+1/6+1/8+1/16+...+1/256+1/512
=(1/2+1/4+1/8+1/16+...+1/256+1/256-1/512)+1/6
=(1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+...+1/128-1/256+1/256-1/512)+1/6
=1-1/512+1/6
=1789/1536
Vậy A=1789/1536
NB
1
28 tháng 8 2019
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)
\(Ax2=2x\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\right)\)
\(Ax2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
\(Ax2-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{128}-\frac{1}{256}\right)\)
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
DT
0
11 tháng 9 2019
Tính \(S=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)
Dùng sai phân như sau
\(2S-S=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)=1-\frac{1}{256}\)
Vậy \(S=1-\frac{1}{256}\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{148}+\frac{1}{256}\)
\(\Rightarrow A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}+\frac{1}{2^8}\)
\(\Rightarrow A=1-\frac{1}{2^7}=1-\frac{1}{128}=\frac{127}{128}\)