\(E=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\) \(E< \frac{3}{4}\)
Những câu hỏi liên quan
Cho E = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.....+\frac{100}{3^{100}}\)
CMR E <\(\frac{3}{4}\)
\(E=\frac{1}{3}+\frac{2}{^{^{3^2}}}+\frac{3}{3^3}+...+\frac{100}{3^{100}}.\)Chứng minh : \(E
CMR : E = \(1-\frac{1}{2^2}-\frac{1}{3^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)
F = \(\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{200^2}< \frac{1}{2}\)
H = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
\(E=1-\frac{1}{2^2}-\frac{1}{3^2}-..........-\frac{1}{2004^2}\)
\(E=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+..........+\frac{1}{2014^2}\right)\)
Ta có : \(E< 1-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{2003.2004}\right)\\ \)
Đặt A= \(1-\left(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2003.2004}\right)\\ =>A=1-\left(1-\frac{1}{2004}\right)\\ =>A=1-\frac{2003}{2004}\\ =>A=\frac{1}{2004}\)
Chắc chắn bạn đã ghi nhầm dấu
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Chứng minh rằng:
\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}<\frac{3}{16}\)
Đang cần rất gấp
Trợ cấp 3 l-i-k-e
Cho \(E=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}.\)\(CMR:E<\frac{3}{4}\)
bài nhà nữa thôi nha
đặt:
\(M=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{4^4}+..+\frac{99}{3^{99}}+\frac{100}{3^{100}}\)
do đó:
\(3M=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{99}{3^{98}}+\frac{100}{3^{99}}\)
=>3M-M=2M=\(1+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)-\frac{100}{3^{100}}\)
ta thấy bthuc trong ngoặc nhỏ hơn 1/2
=>2M<1+1/2
hay M<3/4
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Chứng minh:
E = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.....+\frac{100}{3^{100}}\) < \(\frac{3}{4}\)
=> 3E =1+2/3+3/3^2+...+100/3^99
=> 3E-E=1+1/3+1/3^2+...+1/3^99-100/3^100
=> 2E=1+1/3+1/3^2+...+1/3^99-100/3^100
=> 6E=3+1+1/3+1/3^2+....+1/3^98-100/3^99
=> 6E-2E=3-100/3^99+100/3^100
=> 4E=3-100/3^99+100/3^100
=> E=3/4 -100/3^99.4+100/3^100.4<3/4
Vậy E< 3/4
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Tínha)Aleft(1-frac{1}{1+2}right)left(1-frac{1}{1+2+3}right)...left(1-frac{1}{1+2+3+...+2006}right)b)B1+frac{3}{2^3}+frac{4}{2^4}+...+frac{100}{2^{100}}c)Cfrac{1}{2!}+frac{2}{3!}+...+frac{n-1}{n!}d)D1+2^2+3^2+...+98^2e)E3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1f)F2^{2010}-2^{2009}-...-2-1g)Gleft(frac{1}{4}-1right)left(frac{1}{9}-1right)left(frac{1}{16}-1right)...left(frac{1}{100}-1right)left(frac{1}{121}-1right)
Đọc tiếp
Tính
a)A=\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)
b)\(B=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)
c)C=\(\frac{1}{2!}+\frac{2}{3!}+...+\frac{n-1}{n!}\)
d)D=\(1+2^2+3^2+...+98^2\)
e)E=\(3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
f)F=\(2^{2010}-2^{2009}-...-2-1\)
g)G=\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{100}-1\right)\left(\frac{1}{121}-1\right)\)
câu g)
\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)
\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)
\(=\frac{12}{3}=4\)
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\(E=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
Đặt \(A=\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}\)
\(A=\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+\left(\frac{3}{97}+1\right)+...+\left(\frac{98}{2}+1\right)+1\) ( 99/1 = 99, tất cả 98 ( không tính 99/1) hạng tử trong A đều cộng với 1 , dư ra 1 chỗ cuối)
\(A=\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}+\frac{100}{100}\) ( 100/100=1)
\(A=100.\left(\frac{1}{2}+...+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}\right)\)
Thay A vào E, có:
\(E=\frac{100.\left(\frac{1}{2}+...+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(E=100\)
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\(\Rightarrow E=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+....+\frac{98}{2}+1+1+...+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\) ( Có 99 số 1)
\(\Rightarrow\frac{\frac{1}{99}+1+\frac{2}{98}+\frac{3}{97}+1+...+\frac{98}{2}+1+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)(Nhóm 98 số 1 với 98 phân số đầu ở trên tử)mik viết thiếu nha sorry *-*
\(\Rightarrow E=\frac{\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}+\frac{100}{100}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(\Rightarrow E=\frac{\frac{100}{2}+\frac{100}{3}+\frac{100}{4}+...+\frac{100}{99}+\frac{100}{100}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(\Rightarrow E=\frac{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
\(\Rightarrow E=\frac{100.1}{1}=100\)
~Chúc bạn hok tốt~
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E=\(\frac{1}{3}\)+\(\frac{2}{3^2}\)+\(\frac{3}{3^3}\)+...+\(\frac{100}{3^{100}}\)
CMR:E<\(\frac{3}{4}\)
\(E=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3E=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3E-E=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)
\(2E=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6E=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6E-2E=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4E=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4E=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4E=3-\frac{203}{3^{100}}< 3\)
=> 4E < 3 => E < 3/4
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