Question

Cho tổng gồm 2014 số hạng: \(S=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2014}{4^{2014}}\). Chứng minh rằng \(S< \dfrac{1}{2}\)

Answer 1

4S=1+24+342+....+2014420134S=1+24+342+....+201442013

4S−S=3S=1+24+342+....+201442013−(14+242+343+....+201442014)4S−S=3S=1+24+342+....+201442013−(14+242+343+....+201442014)

3S=1+(24−14)+(342−242)+......+(201442013−201342013)−2014420143S=1+(24−14)+(342−242)+......+(201442013−201342013)−201442014

3S=1+14+142+143+.....+142013−2014420143S=1+14+142+143+.....+142013−201442014

đặt A=1+14+142+143+....+142023A=1+14+142+143+....+142023

4A−A=4+1+14+142+.....+142022−(1+14+142+....+142023)4A−A=4+1+14+142+.....+142022−(1+14+142+....+142023)

3A=4−1420233A=4−142023

A=43−13.42023A=43−13.42023

⇒3S=43−13.42023−201442024⇒3S=43−13.42023−201442024

⇒S=49−19.42023−20143.42024⇒S=49−19.42023−20143.42024

do 49<48=1249<48=12

⇒S=49−19.42023−20143.42024<48=12(đpcm)