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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
S= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
2S= 2(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256)
= 1+1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
=>S = 2S-S =1+1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 -1/2 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64 - 1/128 - 1/256
=1-1/256
=255/256
255/256
ht
và
NHỚ K CHO MIK NHA!!!!!!!!!!!!!!!!!
a) = \(\frac{127}{96}\)
b) = \(\frac{255}{256}\)
c) Mik bỏ nha
d) = \(\frac{1023}{512}\)
e) = \(\frac{2343}{625}\)
1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
= 1 – 1/2 + 1/2- 1/4 + 1/4 – 1/8 + 1/8 – 1/16 + 1/16 – 1/32 + 1/32 – 1/64 + 1/64 – 1/128 + 1/128 – 1/256 – 1/256 – 1/512
= 1 – 1/512
= 511/512
1 a) (x+ 1) + (x + 2 ) + (x + 3) + ... + (x + 100) = 205550 (100 cặp)
=> (x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 205 550
100 số hạng x 100 số hạng
=> 100.x + 100 . 101 : 2 = 205 550
=> 100.x + 5050 = 205 550
=> 100 . x = 205 550 - 5050
=> 100 . x = 200500
=> x = 200500 : 100
=> x = 2005
\(\frac{1999x1999}{1995x1995_{ }}=\frac{1999^2}{1995^2}=\left(\frac{1999}{1995}\right)^2\)\(>1^2\)\(=1\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}+\frac{1}{512}\)
\(A\cdot2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
\(A\cdot2-A=1-\frac{1}{512}\)
\(A=\frac{511}{512}\)
\(2S=2.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{8}+...+\dfrac{1}{256}+\dfrac{1}{384}\right)\)
\(2S=1+\dfrac{2}{3}+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\)
Ta có S = 2S - S
\(\Rightarrow S=1+\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{8}+...+\dfrac{1}{256}-\dfrac{1}{256}-\dfrac{1}{384}\)
\(\Rightarrow S=1+\dfrac{1}{3}-\dfrac{1}{384}=\dfrac{511}{384}\)