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\(A=3^{101}+3^{102}+3^{103}+...+3^{200}\)
\(3A=3^{102}+3^{103}+3^{104}+...+3^{201}\)
\(3A-A=\left(3^{102}+3^{103}+3^{104}+3^{201}\right)-\left(3^{101}+3^{102}+3^{103}+...+3^{201}\right)\)
\(2A=3^{201}-3^{101}\)
\(2A=3^{100}\)
\(\Rightarrow A=3^{100}:2\)
\(A=3^{101}+3^{102}+3^{103}+...+3^{200}\)
\(A=3^{101}+3^{102}+3^{103}+3^{104}+...+3^{197}+3^{198}+3^{199}+3^{200}\)
\(A=3^{100}\left(3+3^2+3^3+3^4\right)+...+3^{196}\left(3+3^2+3^3+3^4\right)\)
\(A=120\left(3^{100}+...+3^{196}\right)⋮120\)
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Ta có:
\(3^{n+2}+3^n-2^{n+2}-2^n\)
= \(3^n.3^2+3^n-2^n.2^2-2^n\)
= \(3^n.\left(3^2+1\right)-2^n.\left(2^2+1\right)\)
= \(3^n.10-2^n.5\)
= \(3^n.10-2^{n-1}.10\) = \(\left(3^n-2^{n-1}\right).10\) => chia hết cho 10
Nhớ cho mk nha bạn !
Ta có 3^n+2+3^n-2^n+2-2^n
=3^n+2+3^n-(2^n+2+2^n)
=3^n.(3^2+1)-2^n.(2^2+1)
=3^n.10-2^n.10
=3^n.10-2^(n-1).10=10.(3^n-2^(n-1)) chia hết cho 10(đpcm)
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1) \(23^{401}+38^{202}-2^{433}=23^{4.100}.23+38^{4.50}.38^2-2^{4.108}.2^1=\left(..1\right).23+\left(..6\right).1444-\left(..6\right).2=\left(..3\right)+\left(..4\right)-\left(..2\right)=\left(..5\right)\)
Chúc bạn có 1 ngày vui vẻ!!!