A=1/31+1/32+...+1/2048 > 3
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11 tháng 6 2020
Đề bài: Tính
\(A=\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\frac{1}{512}+\frac{1}{2048}\)
\(A=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}+\frac{1}{2^{11}}\)
\(2^2.A=2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\)
\(4A-A=\left(2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}+\frac{1}{2^{11}}\right)\)
\(3A=2-\frac{1}{2^{11}}\)
\(\Rightarrow A=\frac{2-\frac{1}{2^{11}}}{3}\)
Vậy \(A=\frac{2-\frac{1}{2^{11}}}{3}\).
11 tháng 6 2020
ta có
A= 1/2+ 1/8+1/32+1/128+1/512+1/2048
=> A= 1/2 +1/ 2^3 +1/2^5 +1/2^7+1/2^9+1/2^11
=> 2^2 A=2+1/2+1/2^3+1/2^5+1/2^7+1/2^9
=> 2^2A-A= (2+1/2+1/2^3+1/2^5+1/2^7+1/2^9)-(1/2+1/2^3+/2^5+1/2^7+1/2^9+1/2^11)
=> 3A= 2- 1/2^11
=>3A= 4095/2048
=> A= 1365/2048
14 tháng 7 2016
A = 1/31 + 1/32 + ... + 1/60
A = (1/31 + 1/32 + ... + 1/40) + (1/41 + 1/42 + ... + 50) + (1/51 + 1/52 + ... + 1/60)
A > 1/40 × 10 + 1/50 × 10 + 1/60 × 10
A > 1/4 + 1/5 + 1/6
A > 1/4 + 1/6 + 1/6
A > 1/4 + 1/3
A > 7/12
LT
3
29 tháng 7 2015
A = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)
Tương tự : (1/41 + 1/42 + ...+ 1/50) < 1/4 ; (1/51 + 1/52+...+1/59+1/60) < 1/5
Mà A = (1/3 + 1/4 + 1/5) = 47/60 > 7/12
Vậy A >7/12
HT
1 tháng 7 2021
Ta có :
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{59}+\frac{1}{60}\)
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(\Rightarrow S>\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)+\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)\)
\(\Rightarrow S>\frac{1}{40}\cdot10+\frac{1}{50}\cdot10+\frac{1}{60}\cdot10\)
\(\Rightarrow S>\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
\(\Rightarrow S>\frac{37}{60}>\frac{36}{60}\) \(=\frac{3}{5}\)
\(\Rightarrow S>\frac{3}{5}\left(đpcm\right)\)
10 tháng 3 2016
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)
Tương tự ta có : (1/41 + 1/42 + ...+ 1/50) > 1/5 ; (1/51 + 1/52+...+1/59+1/60) > 1/6
S > 1/4 + 1/5 + 1/6.
Mà khi đó ta thấy: (1/4 + 1/5 + 1/6) > 3/5
=>S > 3/5 (1)
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Do : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)
=> S < 4/5 (2)
Từ (1) và (2) => 3/5 <S<4/5
A = 1/31 + 1/32 + 1/33 + ... + 1/2048
= (1/31 + 1/32 +...+1/40) + (1/41 + 1/42 +...+1/50) +...+(1/2031 + 1/2032 +..+1/2040)
= 10/40 + 10/50 +...+10/2040
=1/4 + 1/5 +...+1/204
= (1/4 + 1/5 + ... + 1/9) + (1/10 + 1/11 +...+ 1/20) +...+ (1/191 + 1/192 +...+ 1/200)
= (1/4 + 1/5 +...+ 1/9) + 10/20 +...+10/200
= 1/2 + 1/3 + 2(1/4 + 1/5 + .. +1/9) + 1/10 + (1/11 + 1/12+...+ 1/20)
= 5/6 + 2.0,99 + 10/20 > 3
Vậy A > 3 (đpcm)