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Ta có:
\(A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2^2A=1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)
\(\Rightarrow4A=1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)
\(\Rightarrow4A-A=1-\frac{1}{2^{100}}< 1\Rightarrow3A< 1\Rightarrow A< \frac{1}{3}\left(đpcm\right)\)
A=221+241+261+...+21001
\(\Rightarrow 2^{2} A = 1 + \frac{1}{2^{2}} + \frac{1}{2^{4}} + . . . + \frac{1}{2^{98}}\)
\(\Rightarrow 4 A = 1 + \frac{1}{2^{2}} + \frac{1}{2^{4}} + . . . + \frac{1}{2^{98}}\)
\(\Rightarrow 4 A - A = 1 - \frac{1}{2^{100}} < 1 \Rightarrow 3 A < 1 \Rightarrow A < \frac{1}{3} \left(\right. đ p c m \left.\right)\)
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\\ =\left(2-1\right)\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}-\dfrac{1}{2^{99}}\\ =1-\dfrac{1}{2^{99}}< 1\)
Vậy \(B< 1\)
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\)
\(\Rightarrow2B=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\)
\(\Rightarrow2B-B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)
\(\Rightarrow B=1-\dfrac{1}{2^{99}}\)
\(\rightarrow B< 1\rightarrowđpcm\)
`B = 1/2 + (1/2)^2 +... +(1/2)^99`
`=> B = 1/2 + 1/2^2 + ... + 1/2^99`
`=>2B = 1 + 1/2 +... +1/2^98`
`=> 2B - B = (1+1/2 + ... + 1/2^98) -(1/2 + 1/2^2 + ... + 1/2^99)`
`=> B = 1 - 1/2^99<1`
`=> B<1`
Ta có:
\(B=\frac12+\left(\frac12\right)^2+\left(\frac12\right)^3+\cdots+\left(\frac12\right)^{99}\)
\(2B=1+\frac12+\left(\frac12\right)^2+\cdots+\left(\frac12\right)^{98}\)
\(2B-B=\left\lbrack1-\frac12+\left(\frac12\right)^2+\cdots+\left(\frac12\right)^{98}\right\rbrack-\left\lbrack\frac12+\left(\frac12\right)^2+\left(\frac12\right)^3+\cdots+\left(\frac12\right)^{99}\right\rbrack\)
\(B=1-\left(\frac12\right)^{99}\)
⇒ B < 1
Vậy B < 1
Ta có:
\(B=\frac12+\left(\frac12\right)^2+\left(\frac12\right)^3+\cdots+\left(\frac12\right)^{99}\)
\(2B=1+\frac12+\left(\frac12\right)^2+\cdots+\left(\frac12\right)^{98}\)
\(2B-B=\left\lbrack1-\frac12+\left(\frac12\right)^2+\cdots+\left(\frac12\right)^{98}\right\rbrack-\left\lbrack\frac12+\left(\frac12\right)^2+\left(\frac12\right)^3+\cdots+\left(\frac12\right)^{99}\right\rbrack\)
\(B=1-\left(\frac12\right)^{99}\)
⇒ B < 1
Vậy B < 1
Ta có:
$$B = \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + \dots + \left(\frac{1}{2}\right)^{99}$$Nhân đôi hai vế:
$$2B = 1 + \frac{1}{2} + \left(\frac{1}{2}\right)^2 + \dots + \left(\frac{1}{2}\right)^{98}$$Trừ theo vế:
$$2B - B = 1 - \left(\frac{1}{2}\right)^{99}$$$$B = 1 - \left(\frac{1}{2}\right)^{99}$$Vì $\left(\frac{1}{2}\right)^{99} > 0 \Rightarrow B < 1$ (Đpcm).
=b>1 nhé
tick cho mik