cho A=1/2^2+1/2^4+1/2^6+...+1/2^100
chứng minh A<1/3
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ta có :
`1^3` \(⋮\) `1`
\(2^3⋮2\)
\(3^3⋮3\)
.................
\(100^3⋮100\)
`=>` \(1^3+2^3+3^3+...+100^3⋮1+2+3+...+100\)
vậy `A` \(⋮\)`B`
\(\Rightarrow3B=3^2+3^3+3^4+...+3^{101}\\ \Rightarrow3B-B=3^2+3^3+...+3^{101}-3-3^2-3^3-...-3^{100}\\ \Rightarrow2B=3^{101}-3\\ \Rightarrow B=\dfrac{3^{101}-3}{2}\)
B = 31 + 32 + 33 + .... + 399 + 3100
3B = 3(31 + 32 + 33 + ..... + 399 + 3100)
3B = 32 + 33 + 34 +...... + 3100 + 3101
3B - B = 2B = (32 + 33 + 34 + .... + 3100 + 3101) - ( 31 + 32 + 33 + .... + 3100)
2B = (32 - 32) + (33 - 33) +.....+ ( 3100 - 3100) + ( 3101 - 1)
2B = 0 + 0 + 0 + ..... +0 + 3101 - 1
2B = 3101 - 1
B = (3101 - 1) : 2
\(A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\frac{1}{2^8}+...+\frac{1}{2^{100}}\)
\(4A=1+\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{98}}\)
\(3A=4A-A=1-\frac{1}{2^{100}}<1\)
\(A<\frac{1}{3}\)
\(A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\frac{1}{2^8}+...+\frac{1}{2^{100}}\)
\(2^2.A=1+\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{98}}\)
\(2^2.A-A=\left(1+\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\frac{1}{2^8}+...+\frac{1}{2^{100}}\right)\)
\(4.A-A=1-\frac{1}{2^{100}}< 1\)
\(3A< 1\)
\(\Rightarrow A< \frac{1}{3}\left(đpcm\right)\)