Cho S = 1+2+22+23+. . .+22009. So sánh S với 35204
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Có : \(S=1+2+2^2+2^3+....+2^{99}\)
\(\Rightarrow2S=2+2^2+2^3+....+2^{100}\)
\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{100}\right)-\left(1+2+2^2+....+2^{99}\right)\)
\(\Rightarrow S=2^{100}-1< 2^{100}\)
Vậy \(S< 2^{100}\)
S=1+2+22+23+....+299
⇒2S=2+22+23+....+2100
⇒2S−S=2100-1
S=2100-1
vì 2100 -1<2100
⇒S<2100
\(S=1+2+2^2+2^3+...+2^9\)
Đặt \(2S=2+2^2+2^3+2^4+...+2^{10}\)
\(2S-S=2^{10}-1\) hay \(S=2^{10}-1< 2^{10}\)
\(\Rightarrow\) \(2^{10}=2^2.2^8< 5.2^8\)
Vậy \(S< 5.2^8\)
\(#Tuyết\)
2S=2+2^2+...+2^10
=>S=2^10-1=1023
5*2^8=256*5=1280
=>S<5*2^8
S=1+2+22+...+29�=1+2+22+...+29
2S=2(1+2+22+...+210)2�=2(1+2+22+...+210)
2S=2+22+23+...+292�=2+22+23+...+29
2S−S=(2+22+23+...+210)−(1+2+22+...+29)2�−�=(2+22+23+...+210)−(1+2+22+...+29)
\(S=2^{10}-1=2^2.2^8-1=4.2^8-1
HT
S=1+2+22+...+29�=1+2+22+...+29
2S=2(1+2+22+...+210)2�=2(1+2+22+...+210)
2S=2+22+23+...+292�=2+22+23+...+29
2S−S=(2+22+23+...+210)−(1+2+22+...+29)2�−�=(2+22+23+...+210)−(1+2+22+...+29)
\(S=2^{10}-1=2^2.2^8-1=4.2^8-1
Sửa đề: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
Ta có: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
\(=\dfrac{1}{20}+\left(\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{30}\right)+\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)\)
\(\Leftrightarrow S>\dfrac{1}{20}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{4}\)
\(\Leftrightarrow S>\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)(đpcm)
Đặt A=1+2+22+...+220081+2+22+...+22008
=>2A=2.(1+2+22+...+220081+2+22+...+22008)
=>2A=2+22+23+...+220092+22+23+...+22009
=>2A-A=(2+22+23+...+220092+22+23+...+22009)-(1+2+22+...+220081+2+22+...+22008)
=>A=22009−122009−1
=>A=(-1).(−2)2009(−2)2009+(-1).1
=>A=(-1).[(−2)2009+1][(−2)2009+1]
=>A=(-1).(1−22009)(1−22009)
=>1+2+22+...+220081+2+22+...+22008/1-2200922009
=(−1).(1−22009)1−22009(−1).(1−22009)1−22009=-1
Giải:
Đặt A=1+2+22+23+...+22008
2A=2+22+23+24+...+22009
2A-A=(1+2+22+23+...+22008)-(2+22+23+24+...+22009)
A =1-22009
Vậy B=1-22009/1-22009=1
Chúc bạn học tốt!
Bài 1
a) S = 1 + 2 + 2² + 2³ + ... + 2²⁰²³
2S = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²⁴
S = 2S - S = (2 + 2² + 2³ + ... + 2²⁰²⁴) - (1 + 2 + 2² + 2³)
= 2²⁰²⁴ - 1
b) B = 2²⁰²⁴
B - 1 = 2²⁰²⁴ - 1 = S
B = S + 1
Vậy B > S
a,
\(S=1+2+2^2+...+2^{2023}\)
\(2S=2+2^2+2^3+...+2^{2024}\)
\(\Rightarrow S=2^{2024}-1\)
b.
Do \(2^{2024}-1< 2^{2024}\)
\(\Rightarrow S< B\)
2.
\(H=3+3^2+...+3^{2022}\)
\(\Rightarrow3H=3^2+3^3+...+3^{2023}\)
\(\Rightarrow3H-H=3^{2023}-3\)
\(\Rightarrow2H=3^{2023}-3\)
\(\Rightarrow H=\dfrac{3^{2023}-3}{2}\)
mik bt lm câu 1 thôi nha, bn thông cảm:
a = 2007.2009 b = 20082
=(2008 - 1)(2008 + 1)
= 20082 - 1
Ta có, a = 20082 - 1, b = 20082
mà 20082 - 1 < 20082
=> a < b