cho E=1/2^2+1/3^2+1/4^2+...+1/100^2.CMR E<3/4
Những câu hỏi liên quan
Cho E=1/3+2/3^2+3/3^3+...+100/3^100
CMR: E<3/4
1/3E=1/3^2+2/3^3+...+100/3^101
E-1/3E=1/3+1/3^2+1/3^3+...+1/3^100-1/3^101
2/3E=1/3+1/3^2+1/3^3+...+1/3^100-1/3^101
Đặt B=1/3+1/3^2+...+1/3^100
1/3B=1/3^2+1/3^3+...+1/3^101
B-1/3B=1/3-1/3^101
2/3B=1/3-1/3^101
mà 1/3-1/3^101<1/3
=>2/3B<1/3
=>B<1/2
thay B vào E ta có
2/3E=B-1/3^101
Mà B-1/3^101<B
=>2/3E<B
Mà B<1/2
=>2/3E<1/2
=>E<3/4
k cho mk nha
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Cho E = 1/3 + 2/3^2 + 3/3^3 + ... + 100/3^100
CMR: E < 3/4
Cho E=1/3+2/3^2+3/3^3+...+100/3^100
CMR: E<3/4
Cho E=1/3+2/32+3/33+....+100/3100. CMR E<3/4
cho E=1/3 +2/32 +3/33 +...+100/3100 . CMR : E<3/4
Biết \(E=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\). CMR: E < \(\frac{3}{4}\)
Cho E = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.....+\frac{100}{3^{100}}\)
CMR E <\(\frac{3}{4}\)
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= \(1-\left(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2003.2004}\right)\\ =>A=1-\left(1-\frac{1}{2004}\right)\\ =>A=1-\frac{2003}{2004}\\ =>A=\frac{1}{2004}\)
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1)CMR
a) 1/2^2 +1/3^2+......+1/2018^2 < 1
b) 1/2^2 +1/4^2+1/6^2+......+1/2018^2<1/2
c)E=1/2^2+1/3^2+....+1/100^2<3/4
Giúp mình với m đang cần gấp lắm 😱😱😱
a)Ta có: 22>1.2⇒\(\frac{1}{2^2}< \frac{1}{1.2}\)
32>2.3⇒\(\frac{1}{3^2}< \frac{1}{2.3}\)
... 1002>99.100 ⇒ \(\frac{1}{100^2}< \frac{1}{99.100}\)
VT < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)\(=1-\frac{1}{100}< 1\)(ĐPCM)
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