a)\(A-25=350.\left(15^{2007}+15^{2006}+...+15+1\right)\)

\(\frac{A-25}{350}=15^{2007}+15^{2006}+...+15+1\)

\(\frac{\left(A-25\right).15}{350}=15^{2008}+15^{2007}+...+15^2+15\)

\(\Rightarrow\frac{15.\left(A-25\right)}{350}-\frac{A-25}{350}=15^{2008}-1\)

\(\frac{15A-25.15-A+25}{350}=\frac{14A-25.14}{350}=15^{2008}-1\)

\(\frac{14\left(A-25\right)}{350}=15^{2008}-1\)

\(A-25=\frac{350\left(15^{2008}-1\right)}{14}=25.\left(15^{2008}-1\right)\)

\(\Rightarrow A=25.15^{2008}\)

b)15 chia hết cho 5 suy ra 15²⁰⁰⁸ chia hết cho 5²⁰⁰⁸

suy ra 25.15²⁰⁰⁸ chia hết cho 25.5²⁰⁰⁸=5²⁰¹⁰

a)\(A=25.15^{2008}\)

b)A=25.15²⁰⁰⁸ chia hết cho 25.5²⁰⁰⁸=5²⁰¹⁰ ,suy ra điều phải chứng minh

đặt B= 15^2007+15^2006+...+15^2+15+1

  15B=15^2008+15^2007+...+15^3+15^2+15

  15B-B=15^2008-1

  14B=15^2008-1 

   B=(15^2008-1)/14

  thế vào A=350.(15^2008-1)/14+25

   A=25(15^2008-1)+25

  A=25(15^2008-1+1)

   A=25.15^2008 

  A=5^2.5^2008.3^2008

   A=5^2010.3^2008 chia hết cho 5^2010

\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)

\(=4.\left(3+3^3+...+3^{2009}\right)\)

⇒ \(B\) ⋮ 4

b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)

Ta có: \(a^3-25a\)

\(=a^3-a-24a\)

\(=a\left(a^2-1\right)-24a\)

\(=\left(a-1\right)\cdot a\cdot\left(a+1\right)-24a\)

Vì a-1; a và a+1 là ba số nguyên liên tiếp nên \(\left(a-1\right)\cdot a\cdot\left(a+1\right)⋮3\)(1)

Ta có: a-1 và a là hai số nguyên liên tiếp

nên \(\left(a-1\right)\cdot a⋮2\)

\(\Leftrightarrow\left(a-1\right)\cdot a\cdot\left(a+1\right)⋮2\)(2)

mà (2;3)=1(3)

nên từ (1), (2) và (3) suy ra \(\left(a-1\right)\cdot a\cdot\left(a+1\right)⋮6\)

mà \(24a⋮6\)

nên \(\left(a-1\right)\cdot a\cdot\left(a+1\right)-24a⋮6\)

hay \(a^3-25a⋮6\)(đpcm)

Bài 1:

\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)

\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)

Bài 2:

\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)

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