choA=1/2^2+1/4^2+...+1/2018^2. CMR A>1/3
Những câu hỏi liên quan
choA=3^2018-3^2017+3^2016-......-3^3+3^2-3+1
a)rút gọn A
b) tìm chữ số tận cùng của 4*A
CMR: 1/2^2 + 1/3^2 +....+ 1/2018^2 < 3/4
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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choA=2 mũ 0 +2 mũ 1+...+ 2 mũ 2018
B=2 mũ 2019-1
chứng minh rằng A=B
Ta có : \(A=2^0+2^1+2^2+...+2^{2018}\)
\(\Rightarrow2A=2^1+2^2+2^3+...+2^{2019}\)
\(\Rightarrow2A-A=2^{2019}-2^0\)
\(\Rightarrow A=2^{2019}-1\)
\(\Rightarrow A=B\)
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cmr :1/3 + 2/3^2 +........+ 2018/3^2018 < 3/4
A=1/2 +1/2^2+1/2^3+......+1/2^2018
cmr A<1
Ta có :
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2008}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\right)\)
\(A=1-\frac{1}{2^{2018}}< 1\) ( đpcm )
Vậy \(A< 1\)
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CHOA=1.2.3+2.3.5+3.4.7+...=n.(n+1).(2.n+1)
CMR A=n.(n+1)^2.(n+2):2
tra loi giup minh nhanh len ma
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cmr n=1/4^2+1/6^2+1/8^2+...+1/2018^2<1
p= 2!/3!+2!/4!+2!/5!+.+2!/99<1
9 tháng 5 2019 lúc 5:37
CMR: D=\(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+.....+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}< \frac{1}{2}\)
Lời giải:
$D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+......+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}$
$4D=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}$
Trừ theo vế:
\(3D=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow 12D=4+1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2017}}-\frac{2019}{4^{2018}}\)
Trừ theo vế:
$9D=4-\frac{2019}{4^{2018}}+\frac{2019}{4^{2019}}-\frac{1}{4^{2018}}$
$=4-\frac{6061}{4^{2019}}< 4$
$\Rightarrow D< \frac{4}{9}<\frac{4}{8}$ hay $D< \frac{1}{2}$ (đpcm)