Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A=\(\dfrac{7^2-1}{7^4}+\dfrac{7^2-1}{7^8}+...+\dfrac{7^2-1}{7^{100}}=\left(7^2-1\right)\left(\dfrac{1}{7^4}+\dfrac{1}{7^8}+...+\dfrac{1}{7^{100}}\right)=48\cdot B\)Dễ dàng tính được B( nhân hết với 7 mũ 4 roi trừ đi, chia ra là xong) ra đpcm.
Lên lớp 11 thì ta có dạng tổng quát luôn này(tức là nếu n quá lớn thì có thể coi là xảy ra dấu bằng) \(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^n}-\dfrac{1}{7^{n+2}}< \dfrac{1}{50}\)
Đặt \(S=\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{4n-2}}-\dfrac{1}{7^{4n}}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}\)
\(\Rightarrow\dfrac{S}{7^2}=\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}-\dfrac{1}{7^{102}}\)
\(\Rightarrow S+\dfrac{S}{7^2}=\left(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}\right)+\left(\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}-\dfrac{1}{7^{102}}\right)\)
\(\Leftrightarrow\dfrac{50S}{49}=\dfrac{1}{7^2}-\dfrac{1}{7^{102}}< \dfrac{1}{7^2}=\dfrac{1}{49}< \dfrac{1}{50}\)
\(\Leftrightarrow S< \dfrac{1}{50}\)
Vậy \(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}< \dfrac{1}{50}\) (Đpcm)
Đặt :
\(A=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+.........+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)
\(\Leftrightarrow2^2A=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-.......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
\(\Leftrightarrow4A=1-\dfrac{1}{2^2}+\dfrac{1}{2^4}-\dfrac{1}{2^6}+.......-\dfrac{1}{2^{4n-2}}+\dfrac{1}{2^{4n}}-.......-\dfrac{1}{2^{2002}}\)
\(\Leftrightarrow4A+A=\left(1-\dfrac{1}{2^2}+\dfrac{1}{2^4}-\dfrac{1}{2^6}+.......-\dfrac{1}{2^{4n-2}}+\dfrac{1}{2^{4n}}-......-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
\(\Leftrightarrow5A=1-\dfrac{1}{2^{2004}}\)
\(\Leftrightarrow A=\left(1-\dfrac{1}{2^{2004}}\right):5\)
\(\Leftrightarrow A=\dfrac{1}{5}-\dfrac{1}{5}.\dfrac{1}{2^{2004}}< \dfrac{1}{5}=0,2\left(đpcm\right)\)
Đặt \(A=\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{4n-2}}-\dfrac{1}{7^{4n}}+...+\dfrac{1}{7^{98}}+\dfrac{1}{7^{100}}\)
Ta có:
\(\dfrac{A}{7^2}=\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}+\dfrac{1}{7^{102}}\)
\(\Rightarrow A+\dfrac{A}{7^2}=\left(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}+\dfrac{1}{7^{100}}\right)+\left(\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}+\dfrac{1}{7^{102}}\right)\)
\(\Rightarrow\dfrac{50A}{49}=\dfrac{1}{7^2}-\dfrac{1}{7^{102}}< \dfrac{1}{7^2}=\dfrac{1}{49}\)
\(\Rightarrow A< \dfrac{1}{50}\)
=> ĐPCM.
\(\text{Đặt:}S=\frac{1}{7^2}-\frac{1}{7^4}+....-\frac{1}{7^{100}}\Rightarrow49S=1-\frac{1}{7^2}+.....-\frac{1}{7^{98}}\Rightarrow49S+S=50S=\left(1-\frac{1}{7^2}+\frac{1}{7^4}-....-\frac{1}{7^{98}}\right)+\left(\frac{1}{7^2}-\frac{1}{7^4}+....-\frac{1}{7^{100}}\right)=1-\frac{1}{7^{100}}< 1\Rightarrow S< \frac{1}{50}\left(\text{đpcm}\right)\)
\(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)
\(2^2S=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
\(4S=1-\dfrac{1}{2^2}+...+\dfrac{1}{2^{4n}}+\dfrac{1}{2^{4n+2}}+...+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\)
\(4S+S=\left(1-\dfrac{1}{2^2}+...+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
\(5S=1-\dfrac{1}{2^{2004}}< 1\Rightarrow S< \dfrac{1}{5}=0,2\)
1. A = \(\dfrac{3n-7}{n-1}=\dfrac{3n-3}{n-1}+\dfrac{-7}{n-1}=3+\dfrac{-7}{n-1}\)
Tại giá trị \(A\notin Z,3\in Z\)\(\Rightarrow\dfrac{-7}{n-1}\in Z\)\(\Rightarrow n-1\inƯ\left(-7\right)\) với \(x\ne1\) (mẫu sẽ có giá trị là 0 nếu x = 1)
Tại \(n-1=7\)\(\Leftrightarrow n=7+1=8\)
Tại \(n-1=-7\Leftrightarrow n=-7+1=-6\)
Tại \(n-1=1\Leftrightarrow n=1+1=2\)
Tại \(n-1=-1\Leftrightarrow n=-1+1=0\)
2. B = \(\dfrac{4n+1}{2n-3}=\dfrac{4n+6}{2n-3}+\dfrac{-5}{2n-3}=2+\dfrac{-5}{2n-3}\)
Tại giá trị \(B\in Z,2\in Z\)\(\Rightarrow\dfrac{-5}{2n-3}\in Z\)\(\Rightarrow2n-3\inƯ\left(-5\right)\) với \(x\ne\dfrac{3}{2}\)
Tại \(2n-3=5\Leftrightarrow2n=8\Leftrightarrow n=4\)
Tại \(2n-3=-5\Leftrightarrow2n=-2\Leftrightarrow n=-1\)
Tại \(2n-3=1\Leftrightarrow2n=4\Leftrightarrow n=2\)
Tại \(2n-3=-1\Leftrightarrow2n=2\Leftrightarrow n=1\)
a)
\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...........-\dfrac{1}{2004^2}\)
\(\Leftrightarrow B=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..............+\dfrac{1}{2004^2}\right)\)
Đặt :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+.............+\dfrac{1}{2004^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..........................
\(\dfrac{1}{2004^2}< \dfrac{1}{2003.2004}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..............+\dfrac{1}{2003.2004}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{2003}-\dfrac{1}{2004}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2004}\)
\(\Leftrightarrow A< \dfrac{2003}{2004}\)
\(\Leftrightarrow1-A< 1-\dfrac{2003}{2004}\)
\(\Leftrightarrow B< \dfrac{1}{2004}\left(đpcm\right)\)
b) \(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-........+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)
\(\Leftrightarrow2^2S=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.....+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+....+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
\(\Leftrightarrow4S=1-\dfrac{1}{2^2}+.......+\dfrac{1}{2^{4n}}-\dfrac{1}{2^{4n+2}}+.......+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\)
\(\Leftrightarrow4S+S=\left(1-\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)\(\Leftrightarrow5S=1-\dfrac{1}{2^{2004}}< 1\)
\(\Leftrightarrow S< \dfrac{1}{5}=0,2\)
\(\Leftrightarrow S< 0,2\left(đpcm\right)\)
cho mik hỏi mik ko hiểu tại sao từ 1/2^4n-2 khi nhân với 2^2 lại ra đc 1/2^4n vậy? Xin hãy giải đáp giùm mik
Cái này là j
anh dz-À mà ko dz đâu .-. xem giúp em đi