Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
\(S=\left(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\frac{31}{32}+\frac{63}{64}+\frac{127}{128}\right)-6\)
\(S=\left(\frac{64}{128}+\frac{102}{128}+\frac{112}{128}+\frac{120}{128}+\frac{124}{128}+\frac{126}{128}+\frac{127}{128}\right)-6\)
\(S=\frac{64+102+112+120+124+126+127}{128}-6\)
\(S=\frac{775}{128}-6\)
\(S=\frac{775}{128}-\frac{768}{128}\)
\(S=\frac{7}{128}\)
S=1/2+3/4+7/8+15/16+31/32+63/64+127/128 -6
S= 1-1/2 + 1-1/4 + 1-1/8 + 1-1/16 + 1-1/32 + 1-1/64+ 1-1/128 - 6
S= (1+1+1+1+1+1+1-6)- (1/2+1/4+1/8+1/16 + 1/32+1/64+1/128)
S= 1- 111/128
S= 17/128
(Làm lụi nha bn)
Q=\(\dfrac{1}{2}+\left(\dfrac{3}{4}+\dfrac{7}{8}\right)+\left(\dfrac{15}{16}+\dfrac{31}{32}\right)+\left(\dfrac{63}{64}+\dfrac{127}{128}\right)-6\)
Q=\(\dfrac{1}{2}+\dfrac{13}{8}+\dfrac{61}{32}+\dfrac{253}{128}\)\(-6\)
Q= \(\dfrac{64}{128}+\dfrac{208}{128}+\dfrac{244}{128}+\dfrac{253}{128}-6\)
Q= \(\dfrac{769}{128}-6\)
Q=\(\dfrac{769}{128}-\dfrac{768}{128}\)
Q= \(\dfrac{1}{128}\)
Đặt A =\(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+...+\frac{127}{128}-6\)
= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+...+\left(1-\frac{1}{128}\right)-6\)
= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{2^3}\right)+...+\left(1-\frac{1}{2^7}\right)-6\)(7 cặp số)
= \(1-\frac{1}{2}+1-\frac{1}{2^2}+1-\frac{1}{2^3}+...+1-\frac{1}{2^7}-6\)
= \(\left(1+1+1+...+1\right)-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^7}-6\)
= \(1.7-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)-6\)
= \(7-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)-6\)
= \(7-6-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
= \(1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
=> 2A = \(2-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)
Lấy 2A - A = \(\left(2-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\right)-\left(1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\right)\)
A = \(2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^6}-1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)
= \(2-1-1+\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^6}\right)\)
= \(0+\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^3}+...+\frac{1}{2^6}-\frac{1}{2^6}+\frac{1}{2^7}\right)\)
= \(0+\frac{1}{2^7}\)
= \(\frac{1}{2^7}\)
số các chữ số đó là
(200-1):1+1=200
số cặp đó là
200:2=100
tổng 1 cạp là
200+1=201
giá trị bt là
201.100=20100
Ta có : \(S=1+\frac{1}{2}+\frac{1}{4}+......+\frac{1}{1024}\)
\(\Rightarrow2S=2+1+\frac{1}{2}+.....+\frac{1}{1024}\)
\(\Rightarrow2S-S=2-\frac{1}{1024}\)
\(\Rightarrow S=\frac{2047}{1024}\)
\(S=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\\ 3S=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot3\cdot4+...+3\cdot99\cdot100\\ 3S=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\\ 3S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+....+99\cdot100\cdot101-98\cdot99\cdot100\\ 3S=99\cdot100\cdot101\\ S=\dfrac{99\cdot100\cdot101}{3}=33\cdot100\cdot101=3300\cdot101=333300\)
S1 = \(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+...+\frac{127}{128}\)
2S1 = 1 + \(\frac{3}{2}+\frac{7}{4}+\frac{15}{8}+\frac{31}{16}+\frac{63}{32}+\frac{127}{64}\)
2S1 - S1 = S1 = 1 + (1 + 1 + 1 + 1 + 1 + 1) - \(\frac{127}{128}\)= 6 + \(\frac{1}{128}\)
=> S = S1 - 6 = 6 + \(\frac{1}{128}\)- 6 = \(\frac{1}{128}\)
\(S=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\frac{31}{32}+\frac{63}{64}+\frac{127}{128}-6\)
\(S=\frac{1}{2}+\left(\frac{3}{4}+\frac{7}{8}\right)+\left(\frac{15}{16}+\frac{31}{32}\right)+\left(\frac{63}{64}+\frac{127}{128}\right)-6\)
\(S=\frac{1}{2}+\frac{13}{8}+\frac{61}{32}+\frac{253}{128}-6\)
\(S=\frac{64}{128}+\frac{208}{128}+\frac{244}{128}+\frac{253}{128}-6\)
\(S=\frac{769}{128}-6\)
\(S=\frac{769}{128}-\frac{768}{128}\)
\(S=\frac{1}{128}\)
hok tốt!!