Toán (nâng cao)

M

manhnguyen0164

Anh chém tý :D:D:D

a) Ta có A=1+(121+13)+(122+15+16+17)+(123+...+115)+...+(1299+...+121001)A=1+(\dfrac{1}{2^1}+\dfrac{1}{3})+(\dfrac{1}{2^2}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7})+(\dfrac{1}{2^3}+...+\dfrac{1}{15})+...+(\dfrac{1}{2^{99}}+...+\dfrac{1}{2^{100}-1})
A<1+2.12+22.122+...+299.1299A<1+2.\dfrac{1}{2}+2^2.\dfrac{1}{2^2}+...+2^{99}.\dfrac{1}{2^{99}}
=1+1+...+1+1100so^ˊ1=100=\underset{100 số 1}{\underbrace{1+1+...+1+1}}=100
b) Ta có: A=1+12+(13+122)+(15+...+123)+...+(1299+1+...+121001+12100)12100A=1+\dfrac{1}{2}+(\dfrac{1}{3}+\dfrac{1}{2^2})+( \dfrac{1}{5}+...+\dfrac{1}{2^3})+...+(\dfrac{1}{2^{99}+1}+...+\dfrac{1}{2^{100}-1}+\dfrac{1}{2^{100}})-\dfrac{1}{2^{100}}
A>1+12+2.122+22.123+...+299.1210012100A>1+\dfrac{1}{2}+2.\dfrac{1}{2^2}+2^2.\dfrac{1}{2^3}+...+2^{99}.\dfrac{1}{2^{100}}-\dfrac{1}{2^{100}}
=1+12+...+12100so^ˊ12100=5112100>50=1+\underset{100 số}{\underbrace{\dfrac{1}{2}+...+\dfrac{1}{2}}}-\dfrac{1}{2^{100}}=51-\dfrac{1}{2^{100}}>50
 
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