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ta có 1/2mũ 2 +1/3 mũ 2+1/4 mũ 2+...+1/100 mũ 2=1/2.2+1/3.3+1/4.4+...+1/100.100<1/2.3+1/3.4+1/4.5+...+1/99.100+1/100.101=1/2.3-1/100.101=1/6-1/10100=tự tính nhé
\(S=1+2+2^2+...+2^9\)
\(\Rightarrow2S=2+2^2+2^3+...+2^{10}\)
\(\Rightarrow S=2^{10}-1\)
Lại có \(5.2^8=\left(2^2+1\right).2^8=2^{10}+2^8\)
Vậy \(S< 5.2^8\)
Ta có: \(A=2+2^2+2^3+...+2^{100}\)
\(2A=2^2+2^3+2^4+...+2^{101}\)
\(2A-A=2^{101}-2\)
Hay \(A=2^{101}-2\)
Vậy \(A=2^{101}-2\)
_Học tốt_
S=1+2+22+23+...+220
2S=2+22+23+24+...+221
=>S=2S-S=221-1C
Vậy S=221-1
*Ta có: A\(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=\left(2+2^2\right)+2^2\times\left(2+2^2\right)+...+2^{2008}\times\left(2+2^2\right)\)
\(=\left(2+2^2\right)\times\left(1+2^2+2^3+...+2^{2008}\right)\)
\(=6\times\left(2^2+2^3+...+2^{2008}\right)\)
\(=3\times2\times\left(2^2+2^3+...+2^{2008}\right)\)
\(\Rightarrow A⋮3\)
*Ta có: A \(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=2\times\left(1+2+2^2\right)+2^4\times\left(1+2+2^2\right)+...+2^{2008}\times\left(1+2+2^2\right)\)
\(=\left(1+2+2^2\right)\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(=7\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(\Rightarrow A⋮7\)
Mình sửa lại đề C 1 chút xíu
*Ta có: C \(=3^1+3^2+3^3+3^4+...+3^{2010}\)
\(=\left(3+3^2\right)+3^2\times\left(3+3^2\right)+...+3^{2008}\times\left(3+3^2\right)\)
\(=\left(3+3^2\right)\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=12\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=4\times3\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(\Rightarrow C⋮4\)
Các câu khác làm tương tự nhé. Chúc bạn học tốt!
\(S=1+2+2^2+2^3+...+2^{2020}+2^{2021}\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{2020}+2^{2021}\right)\)
\(=3+2^2\left(1+2\right)+...+2^{2020}\left(1+2\right)\)
\(=3+2^2.3+...+2^{2020}.3⋮3\)
VẬY \(S⋮3\)
Trả lời :...........................................
SCSH: (2021 - 1) : 1 = 2020
Tổng: (2021 + 1) : 2 = 1011
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
k nhé
\(S=1+2+2^2+2^3+...+2^{100}\)
\(2S=2+2^2+2^3+2^4+...+2^{101}\)
\(2S-S=\left(2+2^3+..+2^{101}\right)-\left(1+2^2+...+2^{100}\right)\)
\(S=2^{201}-1\)
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