Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 \(A=\frac{2-\frac{1}{2^{11}}}{3}\).
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
Đặt A=1/2+1/4+1/8+1/16+1/32+...+1/2048+1/4096
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{12}}\)
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{11}}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{11}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)\)
\(A=1-\frac{1}{2^{12}}\)
a) \(n^{51}=n\)
\(\Rightarrow n^{51}-n=0\)
\(n\left(n^{50}-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}n=0\\n^{50}-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}n=0\\n^{50}=1\end{cases}}\)
\(\Rightarrow n\in\left\{-1;0;1\right\}\)
b) \(\frac{1}{9}.27^n=3^n\)
\(\Rightarrow3^{-2}.3^{3n}=3^n\)
\(3^{3n-2}=3^n\)
\(\Rightarrow3n-2=n\)
\(3n-n=2\)
\(2n=2\)
\(n=2:2=1\)
c) \(3^{-2}.3^4.3^n=3^7\)
\(3^{n+4-2}=3^7\)
\(3^{n+2}=3^7\)
\(\Rightarrow n+2=7\)
\(\Rightarrow n-7=5\)
d) \(32^{-n}.16^n=2048\)
\(2^{-5n}.2^{4n}=2^{10}\)
\(2^{4n-5n}=2^{10}\)
\(2^{-n}=2^{10}\)
\(\Rightarrow-n=10\)
\(\Rightarrow n=-10\)
Đặt \(A=1+2+4+.........+4096\)
\(2A=2+4+8+......+8192\)
\(\Rightarrow2A-A=8192-1\)
\(\Rightarrow A=8191\)
Đặt \(S=1+2+4+...+1024+2048+4096\)
\(S=1+2^1+2^2+2^3+....+2^{10}+2^{11}+2^{12}\)
\(2S=2+2^2+2^3+....+2^{11}+2^{12}+2^{13}\)
\(2S-S=\left(2+2^2+2^3+....+2^{12}+2^{13}\right)-\left(1+2+2^2+....+2^{11}+2^{12}\right)\)
\(S=2^{13}-1=8192-1=8191\)
Cho A= 1/31+1/32+1/33+.....+1/60
Chứng minh 3/5<A<4/5
GIẢI
Ta có :
\(A=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+.........+\frac{1}{60}\)
\(\Leftrightarrow A=\left(\frac{1}{31}+\frac{1}{32}+....+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+....+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+.....+\frac{1}{60}\right)\left(1\right)\)
Mà:
\(\frac{1}{31}>\frac{1}{32}>\frac{1}{33}>\frac{1}{34}>\frac{1}{35}>\frac{1}{36}>\frac{1}{37}>\frac{1}{38}>\frac{1}{39}>\frac{1}{40}\)
\(\Rightarrow\frac{1}{31}+\frac{1}{32}+........+\frac{1}{40}>\frac{1}{40}+.......+\frac{1}{40}\)
\(\Leftrightarrow\frac{1}{31}+\frac{1}{32}+......+\frac{1}{40}>10\times\frac{1}{40}\)
\(\Leftrightarrow\frac{1}{31}+\frac{1}{32}+..........+\frac{1}{40}>\frac{1}{4}\)
Tương tự:
\(\frac{1}{41}+\frac{1}{42}+.........+\frac{1}{50}>\frac{1}{5}\)
\(\frac{1}{51}+\frac{1}{52}+.....+\frac{1}{60}>\frac{1}{6}\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{36}{60}=\frac{3}{5}\)
Vậy \(\frac{3}{5}< A\left(2\right)\)
Từ (1), ta lại có:
\(\frac{1}{31}+\frac{1}{32}+.......+\frac{1}{40}< 10\times\frac{1}{30}=\frac{1}{3}\)
\(\frac{1}{41}+\frac{1}{42}+..........+\frac{1}{50}< 10\times\frac{1}{40}=\frac{1}{4}\)
\(\frac{1}{51}+\frac{1}{52}+.........+\frac{1}{60}< 10\times\frac{1}{50}=\frac{1}{5}\)
\(\Rightarrow A< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)
Vậy \(A< \frac{4}{5}\left(3\right)\)
Từ (2) và (3) , suy ra:
\(\frac{3}{5}< A< \frac{4}{5}\)
ad ơi cho em hỏi là tại sao lại phải nhóm 10 phân số 1 nhóm vậy ạk
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)