Chứng minh rằng 1/3+1/3^2+1/3^3+...+1/3^2013 < 1/2
Những câu hỏi liên quan
chứng minh rằng 3/1^2.2+5/2^2.3^2+7/3^2.4^2+...+2013/1006^2.1007^2<1
Chứng minh rằng: B= 1/3 + 1/32 + 1/33 +.....+ 1/32012 + 1/32013 < 1/2
Trả lời hộ nha !
3B = 1+1/3+....+1/3^2012
2B=3B-B=(1+1/3+....+1/3^2012)-(1/3+1/3^2+....+1/3^2013) = 1-1/3^2013 < 1
=> B < 1:2 = 1/2
k mk nha
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Cho B = 1 x 2 x 3 x ... x 2012 x (1+1/2+1/3+...+1/2012
Chứng minh rằng B chia hết cho 2013
Chứng minh rằng
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2013\sqrt{2012}}< 2\)
Xét dạng tổng quát: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}.\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\sqrt{n}.\frac{1}{\sqrt{n}}+\sqrt{n}.\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)< \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n+1}}{\sqrt{n+1}}\right)=2.\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Thay vào đề bài ta có:
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2013\sqrt{2012}}\)
\(< 2.\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(< 2.\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2013}}\right)\)
\(< 2.\left(1-\frac{1}{\sqrt{2013}}\right)< 2\left(đpcm\right)\)
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Xét dạng tổng quát :\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}.\frac{1}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)< \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n+1}}{\sqrt{n+1}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Áp dụng vào bài toán:
\(\frac{1}{2\sqrt{1}}< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\right);\frac{1}{3\sqrt{2}}< 2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right);...;\frac{1}{2013\sqrt{2012}}< 2\left(\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
=>\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}...+\frac{1}{2013\sqrt{2012}}< \)\(2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\right)+2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+2\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+...+2\left(\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
mà \(2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\right)+2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+2\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+...+2\left(\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
=\(2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2013}}\right)=2-\frac{2}{\sqrt{2013}}< 2\)
Vậy \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}...+\frac{1}{2013\sqrt{2012}}< 2\)
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Xem thêm câu trả lời
chứng minh rằng:1/5^3+1/6^3+1/7^3+....+1/2013^3<1/40
chứng minh rằng :A = 1/3^2+1/4^2+1/5^2+.....+1/2013^2 <2/5
a/Tính tổng
M=1/5^0+1/5^1+1/5^2+...+1/5^2012
b/Chứng minh rằng 2012^2013-1 và 2012^2013+1 không cùng là số nguyên tố
c/Chứng minh rằng 2+2^2+2^3+...+2^2009+2^2010 chia hết cho 42
a, 5M = 5+1+1/5+1/5^2+.....+1/5^2011
4M=5M-M=(5+1+1/5+1/5^2+.....+1/5^2011)-(1+1/5+1/5^2+.....+1/5^2012)
= 5-1/5^2012
=> M = (5 - 1/5^2012)/4
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Chứng minh rằng 1+2/2+3/2^2+4/2^3+....+2014/2^2013+2015/2^2014 <4
Chứng minh rằng: 1/22 + 1/32 + 1/42 + ........+1/20132 <1