Chứng minh rằng tổng
S = $\frac{1}{2²}$ – $\frac{1}{2^2}$ + $\frac{1}{2^6}$ – …… + $\frac{1}{2^4n-2}$ – $\frac{1}{2^4n}$ + …. + $\frac{1}{2^2002}$ – $\frac{1}{2^2004}$ < 0,2
Chứng minh rằng tổng
S = $\frac{1}{2²}$ – $\frac{1}{2^2}$ + $\frac{1}{2^6}$ – …… + $\frac{1}{2^4n-2}$ – $\frac{1}{2^4n}$ + …. + $\frac{1}{2^2002}$ – $\frac{1}{2^2004}$ < 0,2
Đáp án:
$\rm S<0,2$
Giải thích các bước giải:
`S=1/2^2-1/2^4+1/2^6+…+1/(2^(4n-2))-1/(2^(4n))+…+1/2^2002-1/2^2004`
`2^2 S = 4S = 1/2-1/2^2+1/2^4+…+1/(2^(4n-4))-1/(2^(4n-2)) +…+1/2^2000-1/2^2002`
`4S+S=( 1/2-1/2^2+1/2^4+…+1/(2^(4n-4))-1/(2^(4n-2)) +…+1/2^2000-1/2^2002)+(1/2^2-1/2^4+1/2^6+…+1/(2^(4n-2))-1/(2^(4n))+…+1/2^2002-1/2^2004)`
`5S=1-1/2^2004`
$\rm S=\dfrac{1-\dfrac{1}{2^{2004}} }{5}=\dfrac{1}{5}-\dfrac{1}{2^{2004} . 5}<0,2$
Giải thích các bước giải:
Ta có:
$S=\dfrac1{2^2}-\dfrac{1}{2^4}+\dfrac1{2^6}-…+\dfrac1{2^{4n-2}}-\dfrac1{2^{4n}}+…+\dfrac1{2^{2002}}-\dfrac{1}{2^{2004}}$
$\to 2^2S=1-\dfrac{1}{2^2}+\dfrac1{2^4}-…+\dfrac1{2^{4n-4}}-\dfrac1{2^{4n-2}}+…+\dfrac1{2^{2000}}-\dfrac{1}{2^{2002}}$
$\to S+2^2S=1-\dfrac{1}{2^{2004}}$
$\to 5S=1-\dfrac{1}{2^{2004}}<1$
$\to 5S<1$
$\to S<\dfrac15$
$\to S<0.2$