Tính: 4S-3²⁰²³.Biết S=1-3+3²-3³+...-3²⁰²¹+3²⁰²²
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$S=1-3+3^2-3^3+...-3^{2021}+3^{2022}$
$3S=3-3^2+3^3-3^4+...-3^{2022}+3^{2023}$
$\Rightarrow S+3S=3^{2023}-1$
$\Rightarrow 4S=3^{2023}-1$
$\Rightarrow 4S-3^{2023}=-1$
![](https://rs.olm.vn/images/avt/0.png?1311)
3S=3-3^2+...-3^2022+3^2023
=>4S=3^2023+1
=>4S-3^2023=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có S = 1 + 3 + 32 + ... + 32022
3S = 3 + 32 + 33 + ... + 32023
2S = ( 3 + 32 + 33 + ... + 32023 ) - ( 1 + 3 + 32 + ... + 32022 )
= 32023 - 1
⇒ 4S - 22023 = 2( 32023 - 1 ) - 22023
= 2 . 32023 - 2 - 32023
= 32023( 2 - 1 ) - 2
= 32023 - 2
Vậy 4S = 32023 - 2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1-3+3^2-3^3+...+3^{2008}\)
\(3S=3-3^2+3^3-3^4+...+3^{2009}\)
\(4S=3^{2009}+1\)
\(\Rightarrow4S-1-3^{2009}=3^{2009}+1-1-3^{2009}\)
\(\Rightarrow B=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1+3+3^2+...+3^{2022}\\ 3S=3+3^2+3^3+...+3^{2023}\\ 3S-S=\left(3+3^2+3^3+...+3^{2023}\right)-\left(1+3+3^2+...+3^{2022}\right)\\ 2S=3^{2023}-1\\4S=\dfrac{3^{2023}\times2-1\times2}{2}\\ 4S=\dfrac{\left(3^{2023}-1\right)\times2}{2}\\ 4S=3^{2023}-1\\ 4S-3^{2023}=3^{2023}-1-3^{2023}\\ 4S-3^{2023}=\left(-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1-3+3^2-3^3+....-3^{2007}+3^{2008}\)
\(3S=3-3^2+3^3-3^4+...-3^{2008}+3^{2009}\)
\(4S=3^{2009}+1\)
\(\Rightarrow A=4S-1-3^{2009}\)
\(=\left(3^{2009}+1\right)-1-3^{2009}\)
\(=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
S=1-3+32-33+....+32014-32015
<=> 3S=3(1-3+32-33+....+32014-32015)
<=> 3S=3-32+33-34+....+32015-32016
<=> S+3S=(1-3+32-33+....+32014-32015)+(3-32+33-34+....+32015-32016)
<=> 4S=1-3+32-33+....+32014-32015+3-32+33-34+...+32015-32016
<=> 4S=1-32016
<=> 1-4S=1-1-32016=-32016
=> 2016=5n+1
<=> 5n=2015
<=> n=403
Vậy n=403
![](https://rs.olm.vn/images/avt/0.png?1311)
4S=1*2*3*4+2*3*4(5-1)+......+k*(k+1)(k+2)[(k+3)(k-1)]
tự chứng minh tiếp nhé
\(S=1-3+3^2-3^3+...-3^{2021}+3^{2022}\)
=>\(3S=3-3^2+3^3-3^4+...-3^{2022}+3^{2023}\)
=>\(3S+S=3-3^2+...+3^{2023}+1-3+3^2-...-3^{2021}+3^{2022}\)
=>\(4S=3^{2023}+1\)
=>\(4S-3^{2023}=1\)