S = 1/3 + 1/3² + 1/3³ + ... + 1/3²⁰²¹ + 1/3²⁰²²
⇒ S/3 = 1/3² + 1/3³ + 1/3⁴ + ... + 1/3²⁰²² + 1/3²⁰²³
⇒ 2S/3 = S - S/3
= (1/3 + 1/3² + 1/3³ + ... + 1/3²⁰²¹ + 1/3²⁰²²) - (1/3² +1/3³ + 1/3⁴ + ... + 1/3²⁰²² + 1/3²⁰²³)
= 1/3 - 1/3²⁰²³
⇒ S = (1/3 - 1/3²⁰²³) : 2/3
= (1 - 1/3²⁰²²) : 2
Lại có: 1 - 1/3²⁰²² < 1
⇒ S < 1/2
S= \((\dfrac{1}{2}-\dfrac{1}{2\times3^{2022}})\)
S = 1/3 + 1/32 + 1/33 + ... + 1/32021 + 1/32022
3S = 1 + 1/3 + 1/32 + ... + 1/32020 + 1/32021
3S - S = (1/3 + 1/32 + 1/33 + ... + 1/32021 + 1/32022) - (1 + 1/3 + 1/32 + ... + 1/32020 + 1/32021) = 1 + 1/32022
2S = 1 + 1/32022/2
Vì 1 + 1/32022 > 1 nên là 2S > 1/2
Vậy S > 1/2 (ĐPCM)
S = 1/3 + 1/3² + 1/3³ + ... + 1/3²⁰²¹ + 1/3²⁰²²
⇒ S/3 = 1/3² + 1/3³ + 1/3⁴ + ... + 1/3²⁰²² + 1/3²⁰²³
⇒ 2S/3 = S - S/3
= (1/3 + 1/3² + 1/3³ + ... + 1/3²⁰²¹ + 1/3²⁰²²) - (1/3² +1/3³ + 1/3⁴ + ... + 1/3²⁰²² + 1/3²⁰²³)
= 1/3 - 1/3²⁰²³
⇒ S = (1/3 - 1/3²⁰²³) : 2/3
= (1 - 1/3²⁰²²) : 2
Lại có: 1 - 1/3²⁰²² < 1
⇒ S < 1/2
ta có:
s=1/3+1/32+1/33+1/34+...+1/3^2021 +1/3^2022
Xét dáy số:
T=1/2+1/2^2+1/2^3+...+1/2^2021+1/2^2022
suy ra: 1/3<1/2
1/3^2 < 1/2^3
...
1/3^2021 <1/2^2021
1/3^2022 <1/2^2022
suy ra S<T
ta có:
T=1/2(1+1/2+1/2^2+...+1/2^2020+1/2^2021
=1+1/2+1/2^2+...+1/2^2020+1/2^2021=1/ 1-1/2
vạy
T=1/2 . 2=1
từ S <T và T=1 nên ta có
s<1<1/2
vậy s<1/2
S=1/3+1/32+1/33+...+1/32021+1/32022
=>S/3=1/32+1/33+1/34+...+1/32022+1/32023
=>2S/3=S-S/3
=(1/3+1/32+1/33+...+1/32021+1/32022)-(1/32+1/33+1/34+...+1/32022+1/32023)
=1/3-1/32023
=>S=(1/3-1/32023):2/3
=(1-1/32022):2
Lại có: 1-1/32022<1
=>S<1/2
S = 1/3 + 1/3² + 1/3³ + ... + 1/3²⁰²¹ + 1/3²⁰²²
⇒ S/3 = 1/3² + 1/3³ + 1/3⁴ + ... + 1/3²⁰²² + 1/3²⁰²³
⇒ 2S/3 = S - S/3
= (1/3 + 1/3² + 1/3³ + ... + 1/3²⁰²¹ + 1/3²⁰²²) - (1/3² +1/3³ + 1/3⁴ + ... + 1/3²⁰²² + 1/3²⁰²³)
= 1/3 - 1/3²⁰²³
⇒ S = (1/3 - 1/3²⁰²³) : 2/3
= (1 - 1/3²⁰²²) : 2
Lại có: 1 - 1/3²⁰²² < 1
⇒ S < 1/2
Ta có 3S =1+1/3+1/32+...++3/32021
3S-S=(1+1/3+1/32++1/32021)-(1/3+1/32+1/33+......+1/32021+1/32022)
=1-1/32022
3S-S=1-1/32022
S=1/2-1/2.32022
Vậy S<1/2
