Cho A = 1/3 + 2/3^2 + 3/3^3 + 4/3^4 + …+ 99/3^99 + 100^3^100 cmr A <3/16 (mong các bn giải thích rõ ràng và biệt đội mod trả lời cho em vì 9:00 em đi học ak)
Cho A = 1/3 + 2/3^2 + 3/3^3 + 4/3^4 + …+ 99/3^99 + 100^3^100 cmr A <3/16 (mong các bn giải thích rõ ràng và biệt đội mod trả lời cho em vì 9:00 em đi học ak)
Đặt $A’=\dfrac{1}{3}-\dfrac{2}{3^2}-\dfrac{3}{3^3}-\dfrac{4}{3^4}-…-\dfrac{99}{3^99}-\dfrac{100}{3^{100}}$
$⇒3A=\dfrac{3.1}{3}+\dfrac{2.3}{3^2}+\dfrac{3.3}{3^3}+\dfrac{3.4}{3^4}+…+\dfrac{3.99}{3^{99}}+\dfrac{100.3}{3^{100}}$
$=\dfrac{1}{1}+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+…+\dfrac{99}{3^{98}}+\dfrac{100}{3^{99}}$
⇒$3A-A=\dfrac{1}{1}+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+…+\dfrac{99}{3^{98}}+\dfrac{100}{3^{99}}-\dfrac{1}{3}-\dfrac{2}{3^2}-\dfrac{3}{3^3}-\dfrac{4}{3^4}-…-\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}$
⇒$2A=1+(\dfrac{2}{3}+\dfrac{1}{3})+(\dfrac{3}{3^2}+\dfrac{2}{3^2})+(\dfrac{4}{3^3}-\dfrac{3}{3^3})+(\dfrac{5}{3^4}-\dfrac{4}{3^4})+…+(\dfrac{99}{3^{98}}-\dfrac{98}{3^{98}}+(\dfrac{100}{3^{99}}-\dfrac{99}{3^{99}})-\dfrac{100}{3^{100}}$
$=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+…+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}$
Đặt $B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+…+\dfrac{1}{3^{99}}$
⇒$3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+…+\dfrac{1}{3^{98}}$
⇒$2B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+…+\dfrac{1}{3^{98}}-1-\dfrac{1}{3}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-\dfrac{1}{3^4}-…-\dfrac{1}{3^{99}}$
⇒$2B=1-\dfrac{1}{3^{99}}$
⇒$B=\dfrac{1-\dfrac{1}{3^{99}}}{2}=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}$
Thay vào 2A ta được
$2A=1+\dfrac{1}{2}-\dfrac{1}{2.3^{99}}-\dfrac{100}{3^{100}}$
⇒$A=\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^2.3^{99}}-\dfrac{100}{2^2.3^{100}}<\dfrac{3}{4}$
$A<\dfrac{3}{4}$
o A = 1/3 + 2/3^2 + 3/3^3 + 4/3^4 + …+ 99/3^99 + 100^3^100 cmr A <3/16
đặt A=1/3-2/3^2+3/3^3 -4/3^4+…99/3^99-100/3^100
3A=1-2/3+3/3^2-4/3^3-4/3^3…99/3^98-100/3^99
3A+A=(1-2/3+3/3^2-…+99/3^98-100/3^99)+(1/3-2/3^3-…+99/3^99-100/3^100)
4A=1-1/3+1/3^2-1/3^3-…+1/3^99-100/36100
12A=(3-1+1/3+/1/3^3+…-1/3^98-100/3^100)(3-1+1/3-1/3^2+…-1/3^98-100/3^99)
16A=3-101/3^99-100/3^99-100/3^100<3
suy ra 16A<3
suy ra A<3/16(đpcm)
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