Chứng minh rằng: $\frac{1}{1+3}$ +$\frac{1}{1+3+5}$ +…+$\frac{1}{1+3+…2017}$ <$\frac{3}{4}$ 22/09/2021 Bởi Gabriella Chứng minh rằng: $\frac{1}{1+3}$ +$\frac{1}{1+3+5}$ +…+$\frac{1}{1+3+…2017}$ <$\frac{3}{4}$
Đặt $A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+…+\dfrac{1}{1009^2}$ Ta có: $A<\dfrac{1}{4}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+…..+\dfrac{1}{1008.1009}$ `=>` $A<\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+…+\dfrac{1}{1008}-\dfrac{1}{1009}$ `=>`$A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{1009}=\dfrac{1}{4}+\dfrac{2}{4}-\dfrac{1}{1009}=\dfrac{3}{4}-\dfrac{1}{1009}$ `=>`$A<\dfrac{3}{4}$ Vậy $\dfrac{1}{1+3}+\dfrac{1}{1+3+5}++…+\dfrac{1}{1+3+…+2017}<\dfrac{3}{4}$ Bình luận
Đáp án:
Giải thích các bước giải:
Đặt $A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+…+\dfrac{1}{1009^2}$
Ta có: $A<\dfrac{1}{4}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+…..+\dfrac{1}{1008.1009}$
`=>` $A<\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+…+\dfrac{1}{1008}-\dfrac{1}{1009}$
`=>`$A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{1009}=\dfrac{1}{4}+\dfrac{2}{4}-\dfrac{1}{1009}=\dfrac{3}{4}-\dfrac{1}{1009}$
`=>`$A<\dfrac{3}{4}$
Vậy $\dfrac{1}{1+3}+\dfrac{1}{1+3+5}++…+\dfrac{1}{1+3+…+2017}<\dfrac{3}{4}$