a bằng 1/2+1/2mux2+…+1/2mux2019.so sánh a với 1 14/08/2021 Bởi Brielle a bằng 1/2+1/2mux2+…+1/2mux2019.so sánh a với 1
Đáp án + Giải thích các bước giải: `A=1/2+1/2^2+…+1/2^2019` `=>2A=1+1/2+…+1/2^2018` `=>2A-A=(1+1/2+…+1/2^2018)-(1/2+1/2^2+…+1/2^2019)` `=>A=1-1/2^2019<1` Vậy `A<1` Bình luận
`A= 1/2 + 1/2^2 + …+1/2^2019` `1/2A = 1/2( 1/2 + 1/2^2 +…+1/2^2019)` `1/2A = 1/2^2 + 1/2^3 +…+1/2^2020` `A- 1/2A = 1/2 + 1/2^2 +…+1/2^2019 – 1/2^2 – 1/2^3 -….-1/2^2020` `1/2A = 1/2 – 1/2^2020` `A= (1/2 – 1/2^2020) : 1/2` `A= ( 1/2 – 1/2^2020) .2` `A= 1 – 1/2^2019 < 1` Vậy `A<1` Bình luận
Đáp án + Giải thích các bước giải:
`A=1/2+1/2^2+…+1/2^2019`
`=>2A=1+1/2+…+1/2^2018`
`=>2A-A=(1+1/2+…+1/2^2018)-(1/2+1/2^2+…+1/2^2019)`
`=>A=1-1/2^2019<1`
Vậy `A<1`
`A= 1/2 + 1/2^2 + …+1/2^2019`
`1/2A = 1/2( 1/2 + 1/2^2 +…+1/2^2019)`
`1/2A = 1/2^2 + 1/2^3 +…+1/2^2020`
`A- 1/2A = 1/2 + 1/2^2 +…+1/2^2019 – 1/2^2 – 1/2^3 -….-1/2^2020`
`1/2A = 1/2 – 1/2^2020`
`A= (1/2 – 1/2^2020) : 1/2`
`A= ( 1/2 – 1/2^2020) .2`
`A= 1 – 1/2^2019 < 1`
Vậy `A<1`