Question

[Chocolate ^.^]

Câu 1: Chứng minh:

B= \(5+5^2+5^3+...+5^{2004}\)chia hết cho 126

help nha!!!!!!!

Answer 1

[Đức Hiếu]

\(B=5+5^2+5^3+5^4+5^5+5^6+....+5^{2004}\)

\(B=\left(5+5^2+5^3+5^4+5^5+5^6\right)+...+\left(5^{1999}+5^{2000}+5^{2001}+5^{2002}+5^{2003}+5^{2004}\right)\)

\(B=5.\left(1+5+5^2+5^3+5^4+5^5\right)+....+5^{1999}.\left(1+5+5^2+5^3+5^4+5^5\right)\)

\(B=\left(1+5+5^2+5^3+5^4+5^5\right).\left(5+5^2+....+5^{1999}\right)\)

\(B=3906.\left(5+5^2+....+5^{1999}\right)\)

Vì 3906 chia hết cho 126 nên:

\(3906.\left(5+5^2+....+5^{1999}\right)\) chia hết cho 126

Do đó B chia hết cho 126(đpcm)

Chúc bạn học tốt!!!

Answer 2

[Hoàng Hà Nhi]

Câu 1:

B có 2004 số hạng, ta chia B thành 501 nhóm, mỗi nhóm có 6 số hạng như sau:

\(\)\(B=\left(5+5^2+5^3+5^4+5^5+5^6\right)+....+\left(5^{1999}+5^{2000}+5^{2001}+5^{2002}+5^{2005}+5^{2004}\right)\)

\(B=\left[\left(5+5^4\right)+\left(5^2+5^5\right)+\left(5^3+5^6\right)\right]+....+\left[\left(5^{1999}+5^{2003}\right)+\left(5^{2000}+5^{2003}\right)+\left(5^{2001}+5^{2004}\right)\right]\)

\(B=\left[5\left(1+5^3\right)+5^2\left(1+5^3\right)+5^3\left(1+5^3\right)\right]+...+\left[5^{1999}\left(1+5^3\right)+5^{2000}\left(1+5^3\right)+5^{2001}\left(1+5^3\right)\right]\)

\(B=5.126+5^2.126+5^3.126+...+5^{1999}.126+5^{2000}.126+5^{2001}.126\)\(B=126.\left(5+5^2+5^3+...+5^{1999}+5^{2000}+5^{2001}\right)⋮126\left(đpcm\right)\)

Vậy \(B⋮126\)