Cho A= 1/4+1/4^2+1/4^3+...+1/4^99. Chứng tỏ rằng A<1/3
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\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)
\(=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)
\(\Rightarrow A< 1.\left(\dfrac{1}{2.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(\Rightarrow A< 1+\left(\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Rightarrow A< 1+\left(\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{100}\right)\)
Mà ta thấy \(\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow A< 1+\dfrac{3}{4}=\dfrac{7}{4}\)
\(A=1+4+4^2+4^3+...+4^{99}\)
\(4A=4+4^2+4^3+4^4+...+4^{100}\)
\(4A-A=\left(4+4^2+4^3+4^4+...+4^{100}\right)-\left(1+4+4^2+4^3...+4^{99}\right)\)
\(3A=4^{100}-1\)
\(A=\frac{4^{100}}{3}-\frac{1}{3}=\frac{B}{3}-\frac{1}{3}\)
Vậy \(A< \frac{B}{3}\)
A=1+4+42+...+499
4A=4+42+43+...+4100
4A-A=3A=(4+42+...+4100)-(1+4+42+...+499)
3A=4100-1
Ta thấy: 3A<B =>A<B/3 (điều phải chứng minh)
\(=>4A=4+4^2+...+4^{99}+4^{100}\)
\(=>4A-A=\left(4+4^2+...+4^{99}+4^{100}\right)-\left(1+4+4^2+...+4^{99}\right)\)
\(=>3A=4^{100}-1\)
\(=>A=\frac{4^{100}-1}{3}\)
\(\frac{1}{3}B=\frac{4^{100}}{3}\)
=> A<\(\frac{1}{3}B\)
A = 1 + 4 + 42 + 43 + ... + 499
4A = 4( 1 + 4 + 42 + 43 + ... + 499 )
4A = 4 + 42 + 43 + ... + 4100
4A - A = 3A
= ( 4 + 42 + 43 + ... + 4100 ) - ( 1 + 4 + 42 + 43 + ... + 499 )
= 4 + 42 + 43 + ... + 4100 - 1 - 4 - 42 - 43 - ... - 499
= 4100 - 1
=> \(A=\frac{4^{100}-1}{3}\)
B = 4100 => \(\frac{1}{3}B=4^{100}\cdot\frac{1}{3}=\frac{4^{100}}{3}\)
\(4^{100}-1< 4^{100}\Rightarrow\frac{4^{100}-1}{3}< \frac{4^{100}}{3}\Rightarrow A< \frac{1}{3}B\left(đpcm\right)\)
A = 1/4 + 1/4² + 1/4³ + ... + 1/4⁹⁹
⇒ 4A = 1 + 1/4 + 1/4² + ... + 1/4⁹⁸
⇒ 3A = 4A - A
= (1 + 1/4 + 1/4² + ... + 1/4⁹⁸) - (1/4 + 1/4² + 1/4³ + ... + 1/4⁹⁹)
= 1 - 1/4⁹⁹
⇒ A = (1 - 1/4⁹⁹)/3
Do 1 - 1/4⁹⁹ < 1
⇒ (1 - 1/4⁹⁹)/3 < 1/3
Vậy A < 1/3