Cho S=5+52+53+54+...+52016. Chứng minh S chia hết cho 31
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\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4.\left(3+3^3+...+3^{2009}\right)\)
⇒ \(B\) ⋮ 4
b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)
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S = 5 + 5² + 5³ + 5⁴ + ... + 5²⁰¹²
= (5 + 5² + 5³ + 5⁴) + (5⁵ + 5⁶ + 5⁷ + 5⁸) + ... + (5²⁰⁰⁹ + 5²⁰¹⁰ + 5²⁰¹¹ + 5²⁰¹²)
= 780 + 5⁴.(5 + 5² + 5³ + 5⁴) + ... + 5²⁰⁰⁸.(5 + 5² + 5³ + 5⁴)
= 780 + 5⁴.780 + ... + 5²⁰⁰⁸.780
= 65.12 + 5⁴.65.12 + ... + 5²⁰⁰⁸.65.12
= 65.12(1 + 5⁴ + ... + 5²⁰⁰⁸) ⋮ 65
Vậy S ⋮ 65
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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:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có:
2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100
= 2 + 2 2 + 2 3 + 2 4 + 2 5 +...+ 2 96 + 2 97 + 2 98 + 2 99 + 2 100
= 2. 1 + 2 + 2 2 + 2 3 + 2 4 +...+ 2 96 1 + 2 + 2 2 + 2 3 + 2 4
= 2 . 31 + 2 6 . 31 + . . . + 2 96 . 31
= 2 + 2 6 + . . . + 2 96 . 31 chia hết cho 31
b, Ta có:
5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150
= 5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150
= 5 1 + 5 + 5 3 1 + 5 + 5 5 1 + 5 + . . . + 5 149 1 + 5
= 5 . 6 + 5 3 . 6 + 5 5 . 6 + . . . + 5 149 . 6
= ( 5 + 5 3 + 5 5 + . . . + 5 149 ) . 6 chia hết cho 6
Ta lại có:
5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150
= 5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 +...+ 5 145 + 5 146 + 5 147 + 5 148 + 5 149 + 5 150 (có đúng 25 nhóm)
= [ ( 5 + 5 4 ) + ( 5 2 + 5 5 ) + ( 5 3 + 5 6 ) ] + ... + [ 5 145 + 5 148 ) + ( 5 146 + 5 149 ) + ( 5 147 + 5 150 ]
= [ 5 ( 1 + 5 3 ) + 5 2 ( 1 + 5 3 ) + 5 3 ( 1 + 5 3 ) ] + ... + [ 5 145 1 + 5 3 ) + 5 146 ( 1 + 5 3 ) + 5 147 ( 1 + 5 3 ]
= ( 5 . 126 + 5 2 . 126 + 5 3 . 126 ) + ... + ( 5 145 . 126 + 5 146 . 126 + 5 147 . 126 )
= ( 5 + 5 2 + 5 3 ) . 126 + ( 5 7 + 5 8 + 5 9 ) . 126 + ... + ( 5 145 + 5 146 + 5 147 ) . 126
= 126.[ ( 5 + 5 2 + 5 3 ) + ( 5 7 + 5 8 + 5 9 ) + ... + ( 5 145 + 5 146 + 5 147 ) ] chia hết cho 126.
Vậy 5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150 vừa chia hết cho 6, vừa chia hết cho 126
S=5+52+53+54+...+52016
=(5+52+53)+(54+55+56)+...+(52014+52015+52016)
=5(1+5+52)+54(1+5+52)+...+52014(1+5+52)
=5.31+54.31+...+52014.31
=31(5+54+...+52014)
Vì 31\(⋮\)31 nên 31(5+54+...+52014)
Vậy S \(⋮\) 31
S = 5 + 5 ^ 2 + 5 ^ 3 + 5 ^ 4 + .... + 5 ^ 2016 ( co 2016 số hạng )
S = ( 5 + 5 ^ 2 + 5 ^ 3 ) + ( 5 ^ 4 + 5 ^ 5 + 5 ^ 6) + ..... + ( 5 ^ 2014 + 5 ^ 2015 + 5 ^ 2016 ) Co 2016 : 3 = 672 nhom
S = 5 x ( 1 + 5 + 5 ^ 2 ) + 5 ^ 4 x ( 1 + 5 + 5 ^ 2 ) +...... + 5 ^ 2014 x ( 1 + 5 + 5 ^ 2 )
S = 5 x 31 + 45 ^ 4 x 31 + ... + 5 ^ 2014 x 31
S = ( 5 + 5 ^ 4 + .... + 5 ^ 2014 ) x 31
VÌ 31 chia hết cho 31 nên ( 5 + 5 ^ 4 +.... + 5 ^ 2014 ) x 31 chia hết cho 31, hay B chia hết cho 31