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\(S=1+2+2^2+2^3+...+2^{100}\)
\(2S=2+2^2+2^3+2^4+...+2^{101}\)
\(2S-S=\left(2+2^3+..+2^{101}\right)-\left(1+2^2+...+2^{100}\right)\)
\(S=2^{201}-1\)
Ta có
S = 1 + 2 + 22 + 23 + ....+ 2100
2S = 2 + 22 + 23 + 24 + . ....+ 2101
2S-S = ( 2 + 22 + 23 + 24 + . ....+ 2101) - ( 1 + 2 + 22 + 23 + ....+ 2100)
S = 2 + 22 + 23 + 24 + . ....+ 2101 - 1 -2 - 22 - 23 -....- 2100
S = 2101 - 1
S=1+2+22+23+...+220
2S=2+22+23+24+...+221
=>S=2S-S=221-1C
Vậy S=221-1
lam luon nha
=>S.2=2+2^2+2^3+....+2^2018
=>S.2-S=S=(2+2^2+2^3+...+2^2018)-(1+2+2^2+2^3+...2^2017)
=>S=2^2018-1
2x+2x+1+2x+2+2x+3-480=0
2x+2x.2+2x.22+2x.23=0+480
2x.(1+2+22+23)=480
2x.(1+2+4+8)=480
2x.15=480
2x=480:15
2x=32=25
Vậy x =5
nếu sai thì thông cảm nha
\(S=1+2+2^2+2^3+...+2^{2020}+2^{2021}\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{2020}+2^{2021}\right)\)
\(=3+2^2\left(1+2\right)+...+2^{2020}\left(1+2\right)\)
\(=3+2^2.3+...+2^{2020}.3⋮3\)
VẬY \(S⋮3\)
Trả lời :...........................................
SCSH: (2021 - 1) : 1 = 2020
Tổng: (2021 + 1) : 2 = 1011
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
k nhé
\(S=\frac{3}{2^0}+\frac{3}{2^1}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(2S=6+\frac{3}{2^0}+\frac{3}{2^1}+...+\frac{3}{2^8}\)
2S-S=6-\(\frac{3}{2^9}\)
S=\(5\frac{509}{512}\)
S = 2^0 + 2^2 + 2^4 +...+ 2^100
4S = 2^2 + 2^4 + 2^6 + ... + 2^100 + 2^102
4S - S = 2^2 + 2^4 + 2^6 + ... + 2^100 + 2^102 - ( 2^0 + 2^2 + 2^4 +...+ 2^100 )
3S = 2^102 - 1
S = ( 2^102 - 1 ) / 3
\(S=1+2+2^2+...+2^9\)
\(\Rightarrow2S=2+2^2+2^3+...+2^{10}\)
\(\Rightarrow S=2^{10}-1\)
Lại có \(5.2^8=\left(2^2+1\right).2^8=2^{10}+2^8\)
Vậy \(S< 5.2^8\)
S=1+2+2^2+2^3+...+2^9
2S=2+2^2+2^3+...+2^9+2^10
2S-S=(2+2^2+2^3+...+2^9+2^10)-(1+2+2^2+2^3+...+2^9)
S=2^10-1
5.2^8=(2^2+1).2^8=(2^2.2^8)+(1.2^8)=2^10+2^8
Vì 2^10-1<2^10+2^8=> S<5.2^8
Vậy S < 5. 2^8