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Đặt A = 1 + 5 + 5^2 + 5^3 + 5^4 +...+ 5^402 + 5^403 + 5^404
= (1 + 5 + 5^2) + (5^3 + 5^4 + 5^6) +...+ (5^402 + 5^403 + 5^404)
= (1 + 5 + 5^2) + 5^3(1 + 5 + 5^2) +...+ 5^402(1 + 5 + 5^2)
= 31 + 5^2.31 +...+ 5^402.31
= 31.(1 + 5^2 +... + 5^402) chia hết cho 31.
Vậy A chia hết cho 31 (ĐPCM)
bấm vào đây nhé chung to1 +5+52 +..............+5402+5403+5404 chia het cho 3
gom: (1+5+5^2)+(5^3+5^4+5^5)+....(5^402+5^403+5^404)
=1(1+5+5^2)+5^3(1+5+5^2)+...+5^402(1+5+5^2)
=1.31+5^3.31+...+5^402.31
Vay 1+5+5^2+...+5^403+5^404chia het cho 31
B= 1+ 5+ 5^2+ 5^3+ ... + 5^96+ 5^97+ 5^98
=(1+5+52)+(53+54+55)+....+(596+597+598)
=31+(53.1+53.5+53.52)+....+(596.1+597.5+598.52)
=31+53.(1+5+52)+....+596.(1+5+52)
=31.1+53.31+...+596.31
=31.(1+53+...+596)
=> B chia hết cho 31
B = 1+5+52+53+....+598
B = (1+5+52)+(53+54+55)+....+(596+597+598)
B = 1(1+5+52)+53(1+5+52)+....+596(1+5+52)
B = 1.31 + 53.31+.......+596.31
B = 31.(1+53+.....+596) chia hết cho 31 (đpcm)
B = 5 + 5² + 5³ + ... + 5⁹⁰
= (5 + 5² + 5³) + (5⁴ + 5⁵ + 5⁶) + ... + (5⁸⁸ + 5⁸⁹ + 5⁹⁰)
= 5.(1 + 5 + 5²) + 5⁴.(1 + 5 + 5²) + ... + 5⁸⁸.(1 + 5 + 5²)
= 5.31 + 5⁴.31 + ... + 5⁸⁸.31
= 31.(5 + 5⁴ + ...+ 5⁸⁸) ⋮ 31
Vậy B ⋮ 31
\(B=5+5^2+5^3+...+5^{89}+5^{90}\)
Ta có: \(B=\left(5+5^2+5^3\right)+...+\left(5^{88}+5^{89}+5^{90}\right)\)
\(B=155+...+5^{87}.\left(5+5^2+5^3\right)\)
\(B=155+...+5^{87}.155\)
\(B=155.\left(1+...+5^{87}\right)\)
Vì \(155⋮31\) nên \(155.\left(1+...+5^{87}\right)⋮31\)
Vậy \(B⋮31\)
\(#WendyDang\)
=(1+5+5^2)+...+5^402(1+5+5^2)
=31+...+5^402.31
=31(1+...+5^402) chia hết cho 31
\(1+5+5^2+...+5^{404}=\left(1+5+5^2\right)+...+\left(5^{400}+5^{401}+5^{402}\right)=31+31.5^3+...+31.5^{400}\)
\(=31\left(1+5^3+5^6+...+5^{400}\right)\)chia hết cho 31
\(A=1+5^1+5^2+...+5^{101}\)
\(A=\left(1+5^1+5^2\right)+...+\left(5^{99}+5^{100}+5^{101}\right)\)
\(A=\left(1+5^1+5^2\right)+...+5^{99}.\left(1+5^1+5^2\right)\)
\(A=31+...+5^{99}.31\)
\(A=31.\left(1+...+5^{99}\right)⋮31\left(đpcm\right)\)