5
3
1+1/2+1/3+1/4+...+1/2^100-1
chứng minh rằng 50<A<100
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nếu câu trả lời hữu ích nhé!
2560
1626
Ta có:`A=1/2+1/3+1/4+...+1/(2^100-1)`
`A=1+1/2+(1/3+1/2^2)+(1/5+...+1/2^3)+...+(1/(2^99+1)+...+1/2^100)-1/2^100`
Ta có:
`1+1/2` giữ nguyên
`1/3+1/2^2>1/(2^2).2`
`1/5+...+1/2^3>1/(2^3).2^2`
`...`
`1/(2^99+1)+...+1/2^100>1/(2^100).2^99`
`1/2^100` giữ nguyên
`⇒A>1+1/2+1/(2^2).2+1/(2^3).2^2+...+1/(2^100).2^99`
`⇒A>1+1/(2).100-1/2^100>50(1)`
Lại có:`A=1/2+1/3+1/4+...+1/(2^100-1)`
`⇒A=(1/2+1/3)+(1/4+1/5+1/6+1/7)+(1/8+1/9+...+1/15)+...+(1/2^99+1/(2^99+1)+1/(2^99+2)+...+1/(2^100-1))`
Ta có:
`1/2+1/3<2.(1)/2`
`1/4+1/5+1/6+1/7<4.(1)/4`
`1/8+1/9+...+1/15<8.(1)/8`
`...`
`1/2^99+1/(2^99+1)+1/(2^99+2)+...+1/(2^100-1)<2^99.(1)/2^99`
`⇒A`$\text{<1+2.$\dfrac{1}{2}$+4.$\dfrac{1}{4}$+...+$2^{99}$. $\dfrac{1}{2^{99}}$}$
`⇒A`$\text{<$\underbrace{1+1+1+...+1}_{\text{100 số hạng}}$}$
`⇒A`$\text{<1.100}$
`⇒A`$\text{<100}$`(2)`
Từ `(1)` và `(2)`
`⇒50<A<100`(đpcm)
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