cho \(A=\frac{1}{7^2}-\frac{1}{7^4}+....+\frac{1}{7^{4n-2}}-\frac{1}{7^{4n}}+....+\)\(\frac{1}{7^{100}}\)
CMR:\(\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{4n-2}}-\frac{1}{7^{4n}}+...+\frac{1}{98}-\frac{1}{100}< \frac{1}{50}\)
Chứng minh rằng : \(\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{4n-2}}-\frac{1}{7^{4n}}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}< \frac{1}{50}\)
M = 512 - 512/2 - .... - 512/2^10
= 2^9 - 2^9 / 2 - 2^9/2^2 - ...2^9/2^10
= 2^9 - 2^8 - 2^7 - 2^6 -.... - 1/2
2M = 2^10 - 2^9 - 2^8 - .... - 1
2M - M = 2^10 - 2^9 - 2^8 -... -1 - 2^9 + 2^8 + 2^7 +... + 1 + 1/2
M = 2^10 - 2.2^9 + 1/2
M = 2^10 - 2^10 + 1/2
M = 1/2
Đặt \(A=\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{4n-2}}-\frac{1}{7^{4n}}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
\(\Rightarrow49A=1-\frac{1}{7^2}+...+\frac{1}{7^{4n-4}}-\frac{1}{7^{4n}}+..+\frac{1}{7^{96}}-\frac{1}{7^{98}}\)
\(\Rightarrow49A+A=50A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{50}=\frac{1}{50}-\frac{1}{7^{100}.50}< \frac{1}{50}\left(ĐPCM\right)\)
Chứng minh \(\frac{1}{7^2}-\frac{1}{74}+...+\frac{1}{7^{4n-2}}-\frac{1}{7^{4n}}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}<\frac{1}{50}\)
Chứng minh:
\(\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{4n-2}+\frac{1}{4n}+...+\frac{1}{98}+\frac{1}{100}<\frac{1}{50}\)
đề có thiếu hay thừa gì ko nhỉ? tại cái này hình như vế trái gồm 2 dãy quy luật.dãy có các số hạng là bội của 1/7 ko thấy số cuối =="
Biểu thức ko có quy luật
=> sai đề
=> bỏ :V
Chứng minh rằng: \(\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{4n-2}}-\frac{1}{7^{4n}}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}<\frac{1}{50}\)???
Bạn Nào giỏi thì giúp mik với nhé? Mik đang cần gấp. thanks.
\(A=\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
\(\Rightarrow7^2.A=\frac{1}{1}-\frac{1}{7^2}+...+\frac{1}{7^{96}}-\frac{1}{7^{98}}\)
\(\Rightarrow49A+A=1-\frac{1}{7^{100}}\)
\(50A=1-\frac{1}{7^{100}}
CMR : \(\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{4n-2}}-\frac{1}{7^n}+...+\frac{1}{7^{98}}+\frac{1}{7^{100}}< \frac{1}{50}\)
Giải hộ mình nhé
Đặt \(A=\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}+\frac{1}{7^8}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
Nhân \(\frac{1}{7^2}\)vào A. Ta được:
\(A.\frac{1}{7^2}=\frac{1}{7^4}-\frac{1}{7^6}+\frac{1}{7^8}-...-\frac{1}{7^{98}}+\frac{1}{7^{100}}+\frac{1}{7^{102}}\)
\(A=\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-\frac{1}{7^8}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
Ta có: \(\frac{1}{7^2}.A+A=\frac{1}{49}-\frac{1}{7^{102}}\Rightarrow\frac{50}{49}.A=\frac{1}{49}-\frac{1}{7^{102}}\)
\(\Rightarrow A=\left(\frac{1}{49}-\frac{1}{7^{102}}\right)\frac{49}{50}< \frac{1}{5}^{\left(đpcm\right)}\)
Chứng minh rằng :
\(\frac{1}{7^2}\)- \(\frac{1}{7^4}\)+........+ \(\frac{1}{7^{4n-2}}\)- \(\frac{1}{7^{4n}}\) +.......+\(\frac{1}{7^{98}}\)- \(\frac{1}{7^{100}}\)< \(\frac{1}{50}\)
\(A=\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
\(7^2.A=1-\frac{1}{7^2}+\frac{1}{7^4}-...+\frac{1}{7^{100}}-\frac{1}{7^{102}}\)
\(\Rightarrow49A+A=1-\frac{1}{7^{102}}
Ta đặt : A = 1/7 2 - 1/7 4 + ... + 1/7 9s - 1/7 100
=> : A = 1 - 1/7 2 + 1/7 4 -... + 1/7 100 - 1/7 102
=< : 49 + 4 = 1 - 1/7 102 < 1
<=> : 50A < 1 => 1/50
mk biết rõ lun
a, có bao nhiêu số tự nhiên nhỏ hơn 2016 thỏa mãn ko chia hết cho 7.vì sao?
b, cho A =\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...+\frac{1}{15}\)
Chứng tỏ A < 2
c,tìm số nguyên n để 3 - 4n chia hết n + 5
CMR:Với mọi số tự nhiên n \(\ne\)0 ta đều có:
a.\(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+...+\frac{1}{\left(3n-1\right)\times\left(3n+2\right)}=\frac{1}{6n+4}\)
b.\(\frac{5}{3\times7}+\frac{5}{7\times11}+\frac{5}{11\times15}+...+\frac{5}{\left(4n-1\right)\times\left(4n+3\right)}=\frac{5n}{4n+3}\)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)
\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)
\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)
b)\(VT=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left[\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{1}{3}-\frac{1}{4n+3}\right]=\frac{5}{4}\cdot\left[\frac{4n+3}{3\left(4n+3\right)}-\frac{3}{3\left(4n+3\right)}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{4n+3-3}{12n+9}\right]\)\(=\frac{5}{4}\cdot\frac{4n}{12n+9}=\frac{5n}{12n+9}\)