B=\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{2011}{2012!}\)
Chứng minh rằng D=\(\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+....+\frac{2!}{100!}< 1\)
Câc bạn giúp mik với nha!,cảm ơn nhìu
(haiz,làm thì ít mà hỏi thì nhìu)
Những câu hỏi liên quan
Tính nhanh
\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+\frac{2011}{4}+\frac{2010}{5}+...+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}}\)
Giải tự luận hộ mình nha!!!!!!!! Mình cảm ơn!!!
Đặt phân thức trên là D
=> D=(1+1+1+1+...+1+2013/2+2012/3+...+2/2013+1/2014)/(1/2+1/3+1/4+...+1/2014)
=> D=(1+2013/2+1+2012/3+1+2011/4+...+1+2/2013+1+1/2014+1)/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=(2015/2+2015/3+2015/4+...+2015/2013+2015/2014+1)/(1/2+1/3+1/4+...+1/2014)
=> D=[2015*(1/2+1/3+1/4+1/5+....+1/2014)]/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=2015
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UwU
ư uwsuuuuuuuuuuuu kimochiiiiiiiiiiiiiiiiiiii
đùa thôi đáp án: 2015 nha bn
ư ư wsuuuuuuuuuuuuuuuuuuuuuuuuuu kimmmmmooooochiiiiiiiiiii
À quên nhớ FOLOW CHO TUI NHA!
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1.Chứng minh rằng: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^3.4^2}+...+\frac{19}{9^2.10^2}< 1\)
2.Chứng minh rằng: \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
Làm nhanh giúp mình nhé mọi người !!!
Bài 1:
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
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Có phải ở sách NCPT ko bn
Bài 2: Đặt \(B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(3B=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3B-B=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)
\(2B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6B=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6B-2B=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4B=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{203}{3^{100}}< 3\)
\(B< \frac{3}{4}\left(đpcm\right)\)
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CÁC PẠN ƠI GIÚP MIK VỚI LÀM ƠN......
Chứng minh rằng: \(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}< \frac{1}{10}\)
1/ Chứng tỏ rằng frac{1}{1cdot2}+frac{1}{2cdot3}+frac{1}{3cdot4}+...+frac{1}{99cdot100}12/ Chứng tỏ rằng Bfrac{1}{4}+frac{1}{5}+frac{1}{6}+...+frac{1}{19}13/ Rút gọn biểu thức A1+frac{1}{2}+frac{1}{2^2}+frac{1}{2^3}+...+frac{1}{2^{2012}}4/ Tính nhanhfrac{frac{4}{2010}+frac{4}{2011}-frac{4}{2012}}{frac{5}{2010}+frac{5}{2011}-frac{5}{2012}}-frac{frac{1}{123}-frac{1}{19}+frac{1}{371}-frac{1}{5}}{-frac{5}{123}+frac{5}{19}-frac{5}{371}+1} GIÚP ĐƯỢC CÂU NÀO THÌ GIÚP NHÉ, MÌNH TICK CHO
Đọc tiếp
1/ Chứng tỏ rằng \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}<1\)
2/ Chứng tỏ rằng \(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}<1\)
3/ Rút gọn biểu thức \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
4/ Tính nhanh\(\frac{\frac{4}{2010}+\frac{4}{2011}-\frac{4}{2012}}{\frac{5}{2010}+\frac{5}{2011}-\frac{5}{2012}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)
GIÚP ĐƯỢC CÂU NÀO THÌ GIÚP NHÉ, MÌNH TICK CHO
c)\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
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1/
A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1/1-1/100
Vì 1/100>0
-->1/1-1/100<1
-->A<1
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a)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)=\(\frac{99}{100}<1\)
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Chứng minh rằng: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}\)< 1
Ta có : \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};\frac{1}{4^2}<\frac{1}{3.4};...;\frac{1}{2011^2}<\frac{1}{2010.2011};\frac{1}{2012^2}<\frac{1}{2011.2012}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2010+2011}+\frac{1}{2011.2012}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2012}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}<\frac{1}{1}-\frac{1}{2012}\)
Vì \(\frac{1}{2012}>0\) => \(\frac{1}{1}-\frac{1}{2012}<1\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}<1\)
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zee that tri tuệ ve toan day so; ok 10đ
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Chứng minh rằng: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}< 1\)
Ta có:\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};..........;\frac{1}{2012^2}< \frac{1}{2011.2012}\)
Nên \(\frac{1}{2^2}+\frac{1}{3^2}+........+\frac{1}{2012^2}< \frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{2011.2012}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{2011}-\frac{1}{2012}\)
\(=1-\frac{1}{2012}< 1\)
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ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};\frac{1}{5^2}< \frac{1}{4.5};...;\frac{1}{2011^2}< \frac{1}{2010.2011};\)\(\frac{1}{2012^2}< \frac{1}{2011.2012}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2010.2011}+\frac{1}{2011.2012}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2010}-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2012}\)
\(=1-\frac{1}{2012}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}< 1\left(đpcm\right)\)
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Ta thấy \(\frac{1}{2^2}< \frac{1}{1.2}\);\(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2012^2}< \frac{1}{2011.2012}\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+....+\frac{1}{2011^2}+\frac{1}{2012^2}\)
Suy ra \(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2011.2012}=B\)
Ta có \(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2011.2012}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2011}-\frac{1}{2012}\)
\(B=1-\frac{1}{2012}< 1\)
Vì \(B< 1\)mà \(A< B\)nên \(A< 1\)
Khi đó \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+....+\frac{1}{2011^2}+\frac{1}{2012^2}< 1\)
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chứng minh rằng
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}\)<1
Ta có
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.....;\frac{1}{2012^2}< \frac{1}{2011.2012}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2012^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2011.2012}\)
= 1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2011}-\frac{1}{2012}\)
=1-\(\frac{1}{2012}\)=\(\frac{2011}{2012}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2012^2}< 1\)
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Chứng minh rằng: A=\(\frac{1}{2}\)-\(\frac{2}{^{2^2}}\)+\(\frac{3}{2^3}\)-\(\frac{4}{2^4}\)+.....+\(\frac{99}{2^{99}}\)-\(\frac{100}{2^{100}}\)<\(\frac{2}{9}\)
giúp mik nha. Mik cần gấp nha, ai giúp mik kết bạn và like nha
giúp mình làm bài này nha : )))
Chứng minh rằng
\(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)