Toán xét tổng T=2/2+3/2^2+4/2^3+…+2019/2^2018+2020/2^2019 so sánh T với 3 13/09/2021 By Iris xét tổng T=2/2+3/2^2+4/2^3+…+2019/2^2018+2020/2^2019 so sánh T với 3
Đáp án: $T<3$ Giải thích các bước giải: Ta có: $T=\dfrac22+\dfrac3{2^2}+\dfrac4{2^3}+…+\dfrac{2019}{2^{2018}}+\dfrac{2020}{2^{2019}}$ $\to 2T=2+\dfrac3{2}+\dfrac4{2^2}+…+\dfrac{2019}{2^{2017}}+\dfrac{2020}{2^{2018}}$ $\to 2T-T=2+\dfrac12+\dfrac1{2^2}+…+\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$ $\to T=2+\dfrac12+\dfrac1{2^2}+…+\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$ Mà $A=\dfrac12+\dfrac1{2^2}+…+\dfrac{1}{2^{2018}}$ $\to 2A=1+\dfrac12+…+\dfrac1{2^{2017}}$ $\to 2A-A=1-\dfrac{1}{2^{2018}}$ $\to A=1-\dfrac{1}{2^{2018}}$ $\to T=2+1-\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$ $\to T=3-\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$ $\to T<3$ Trả lời
Đáp án: $T<3$
Giải thích các bước giải:
Ta có:
$T=\dfrac22+\dfrac3{2^2}+\dfrac4{2^3}+…+\dfrac{2019}{2^{2018}}+\dfrac{2020}{2^{2019}}$
$\to 2T=2+\dfrac3{2}+\dfrac4{2^2}+…+\dfrac{2019}{2^{2017}}+\dfrac{2020}{2^{2018}}$
$\to 2T-T=2+\dfrac12+\dfrac1{2^2}+…+\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$
$\to T=2+\dfrac12+\dfrac1{2^2}+…+\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$
Mà $A=\dfrac12+\dfrac1{2^2}+…+\dfrac{1}{2^{2018}}$
$\to 2A=1+\dfrac12+…+\dfrac1{2^{2017}}$
$\to 2A-A=1-\dfrac{1}{2^{2018}}$
$\to A=1-\dfrac{1}{2^{2018}}$
$\to T=2+1-\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$
$\to T=3-\dfrac{1}{2^{2018}}-\dfrac{2020}{2^{2019}}$
$\to T<3$