Question

CMR

( 11 + 11² + 11³ +...+ 11^7 + 11^8) chia hết cho 12

Answer 1

\(11+11^2+11^3+11^4+11^5+11^6+11^7+11^8\)

\(=11\left(1+11\right)+11^3\left(1+11\right)+11^5\left(1+11\right)+11^7\left(1+11\right)\)

\(=\left(11+11^3+11^5+11^7\right).12⋮12\)

Vậy ...

Answer 2

Đặt A=\(11+11^2+11^3+....+11^7+11^8\)

\(\Leftrightarrow A=\left(11+11^2\right)+\left(11^3+11^4\right)+...+\left(11^7+11^8\right)\)

\(\Leftrightarrow A=11\left(1+11\right)+11^3\left(1+11\right)+....+11^7\left(1+11\right)\)

\(\Leftrightarrow A=11\cdot12+11^3\cdot12+...+11^7\cdot12\)

\(\Leftrightarrow A=12\left(11+11^3+....+11^7\right)\)

=> A chia hết cho 12 (đpcm)

Answer 3

ta có \(\left(11+11^2+11^3+...+11^7+11^8\right)=\left(11+11^2\right)+\left(11^3+11^4\right)+...+\left(11^7+11^8\right)\)

=\(11\left(1+11\right)+11^3\left(1+11\right)+...+11^7\left(1+11\right)\)

= \(12\left(11+11^3+..+11^7\right)\)\

=> \(\left(11+11^2+...+11^8\right)⋮12\)

Answer 4

gọi M= như trên

=> M=\(11\left(1+11\right)+11^3\left(1+11\right)+...+11^7\left(1+11\right)\)

=> \(M=12\left(11+11^3+...+11^7\right)⋮12\)(dpcm)

Answer 5

Đặt \(A=11+11^2+11^3+...+11^7+11^8\)

Ta có: \(A=11+11^2+11^3+...+11^7+11^8\)

\(\Rightarrow A=\left(11+11^2\right)+\left(11^3+11^4\right)+...+\left(11^7+11^8\right)\)

\(\Rightarrow A=132+11^2\left(11+11^2\right)+...+11^6\left(11+11^2\right)\)

\(\Rightarrow A=132\left(1+11^2+...+11^6\right)\)

Vì \(132⋮12\Rightarrow132\left(1+11^2+...+11^6\right)⋮12\)

\(\Rightarrow11+11^2+11^3+...+11^8⋮12\)

\(\Rightarrow A⋮12\)

\(\Rightarrowđpcm\)

Answer 6

Có 11 + 11² + 11³ +...+ 11⁷ + 11⁸

=(11+11²)+(11³+11⁴)+....+(11⁷+11⁸)

= 132+11²(11+11²)+....+11⁶(11+11²)

= 1.132+11².132+....+11⁶ . 132

=132.(1+11²+....+11⁶)

Vì 132 chia hết cho 12 nên....

Bn tự làm phần cuối nha

Answer 7

Ta có:11+11²+11³+...+11⁷+11⁸

        =(11+11²)+(11³+11⁴)+...+(11⁷+11⁸)

        =11(1+11)+11²(1+11)+...+11⁶(1+11)

        =12(11+11²+...+11⁶) chia hết cho 12

Answer 8

11 + 11² + 11³ +...+ 11⁷ + 11⁸  \(⋮\)12

Ta có : ( 11 + 11²) + (11³ + 11⁴ ) + .....+ (11⁷ + 11⁸)

       =  ( 11 + 11²) + 11^3.  ( 11 + 11²) + ........ + 11⁷ +  ( 11 + 11²)

       =   132      + 11³ . 132 + ..... + 11⁷ + 132

       = 132 . ( 11³ + ...... + 11⁷ ) \(⋮\)12 \(\rightarrowĐPCM\)

# HOK TỐT #

Answer 9

A= 11+11^+2+11^3+11^4+  .......+11^7+11^8

=(11+11^2)+(11^3+11^4)+..........+(11^17+11^18)

do (11+11^2)\(⋮\)12,  (11^3+11^4)\(⋮\)12 ,......,(11^17+11^8)\(⋮\)12

=>Achia het cho 12