CMR A= 1/3 + 2/3^2 + 3/3^3 +…………………..+100/3^99 <3/4 MONG CÁC ANH CHỊ GIÚP EM NHANH NHA ! 23/08/2021 Bởi Maya CMR A= 1/3 + 2/3^2 + 3/3^3 +…………………..+100/3^99 <3/4 MONG CÁC ANH CHỊ GIÚP EM NHANH NHA !
Giải thích các bước giải: Ta có: $A=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+…+\dfrac{100}{3^{99}}$ $\to \dfrac13A=\dfrac{1}{3^2}+\dfrac{2}{3^3}+\dfrac{3}{3^4}+…+\dfrac{100}{3^{100}}$ $\to A-\dfrac13A=\dfrac13+\dfrac1{3^2}+\dfrac{1}{3^3}+…+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}$ $\to \dfrac23A=\dfrac13+\dfrac1{3^2}+\dfrac{1}{3^3}+…+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}$ Lại có : $B=\dfrac13+\dfrac1{3^2}+\dfrac{1}{3^3}+…+\dfrac{1}{3^{99}}$ $\to 3B=1+\dfrac13+\dfrac1{3^2}+…+\dfrac{1}{3^{98}}$ $\to 3B-B=1-\dfrac{1}{3^{99}}$ $\to 2B=1-\dfrac{1}{3^{99}}$ $\to B=\dfrac12(1-\dfrac{1}{3^{99}})$ $\to \dfrac23A=\dfrac12(1-\dfrac{1}{3^{99}})-\dfrac{100}{3^{100}}$ $\to \dfrac23A=\dfrac12-\dfrac{1}{2\cdot 3^{99}}-\dfrac{100}{3^{100}}$ $\to \dfrac23A<\dfrac12$ $\to A<\dfrac34$ Bình luận
Giải thích các bước giải:
Ta có:
$A=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+…+\dfrac{100}{3^{99}}$
$\to \dfrac13A=\dfrac{1}{3^2}+\dfrac{2}{3^3}+\dfrac{3}{3^4}+…+\dfrac{100}{3^{100}}$
$\to A-\dfrac13A=\dfrac13+\dfrac1{3^2}+\dfrac{1}{3^3}+…+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}$
$\to \dfrac23A=\dfrac13+\dfrac1{3^2}+\dfrac{1}{3^3}+…+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}$
Lại có :
$B=\dfrac13+\dfrac1{3^2}+\dfrac{1}{3^3}+…+\dfrac{1}{3^{99}}$
$\to 3B=1+\dfrac13+\dfrac1{3^2}+…+\dfrac{1}{3^{98}}$
$\to 3B-B=1-\dfrac{1}{3^{99}}$
$\to 2B=1-\dfrac{1}{3^{99}}$
$\to B=\dfrac12(1-\dfrac{1}{3^{99}})$
$\to \dfrac23A=\dfrac12(1-\dfrac{1}{3^{99}})-\dfrac{100}{3^{100}}$
$\to \dfrac23A=\dfrac12-\dfrac{1}{2\cdot 3^{99}}-\dfrac{100}{3^{100}}$
$\to \dfrac23A<\dfrac12$
$\to A<\dfrac34$