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a) Ta có:
\(A=2+2^2+2^3+...+2^{24}\)
\(\Rightarrow A=\left(2+2^2+2^3\right)+...+\left(2^{22}+2^{23}+2^{24}\right)\)
\(\Rightarrow A=14+...+2^{21}.\left(2+2^2+2^3\right)\)
\(\Rightarrow A=14+...+2^{21}.14\)
\(\Rightarrow A=\left(1+...+2^{21}\right).14⋮14\)( đpcm )
\(A=2+2^2+2^3+...+2^{24}\)
\(\Rightarrow A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{21}+2^{22}+2^{23}+2^{24}\right)\)
\(\Rightarrow A=2\left(1+2+2^2+2^3\right)+...+2^{21}\left(1+2+2^2+2^3\right)\)
\(\Rightarrow A=2.15+...+2^{21}.15\)
\(\Rightarrow A=15\left(2+...+2^{21}\right)⋮15\left(đpcm\right)\)
b) Mk sửa đề chút là A chia 16 dư 15 nhé
Ta có:
\(A=2+2^2+2^3+...+2^{24}\)
\(\Rightarrow A=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{20}+2^{21}+2^{22}+2^{23}+2^{24}\right)\)
\(\Rightarrow A=2\left(1+2+2^2+2^3+2^4\right)+...+2^{20}\left(1+2+2^2+2^3+2^4\right)\)
\(\Rightarrow A=2.31+...+2^{20}.31\)
\(\Rightarrow A=\left(2+2^{20}\right).31\)
Vì 31 chia 16 dư 15 nên suy ra đpcm
Ta có:
A=(\(2+2^2+2^3+2^4\))+....+(\(2^{21}+2^{22}+2^{23}+2^{24}\))
A=2(1+2+\(2^2+2^3\))+....+\(2^{21}\)(\(1+2+2^2+2^3\))
A=2.15+....+\(2^{21}.15\)
A=15(2+\(2^5+...+2^{21}\))
nên A chia hết cho 15.
\(A=3^2.3^{k+1}+3^{k+1}+2^2.2^{k+1}+2.2^{k+1}\)\(=3^{k+1}\left(3^3+1\right)+2^{k+1}\left(2^2+2\right)\)
\(A=28.3^{k+1}+6.2^{k+1}\)\(=6.\left(14.3^k+2^{k+1}\right)\) chia hết cho 6
3k+3 +3k+1+2k+3+2k+2=3k.9+3k.3+2k.8+2k.4=3k.12+2k.12=(3k+2K)12 chia het 6
b) dễ lắm cậu tự làm nha , tách ra thành 2 vế rồi rút gọn lại
c) \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n.9-2^n.4+3^n.1-2^n.1\)
\(=3^n.\left(9+1\right)-2^n.\left(4+1\right)\)
\(=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.2.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10.\left(3^n.2^{n-1}\right)\)