Chứng minh rằng :
K= 10^n+72n-1 chia hết cho 81
K
Khách
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31 tháng 3 2016
10^n+72n-1
=10^n-1+72n
=(10-1)[10^(n-1)+10^(n-2)+...+10+1]+72n
=9[10^(n-1)+10^(n-2)+...+10+1]-9n+81n
=9[10^(n-1)+10^(n-2)+...+10+1-n]+81n
=9[(10^(n-1)-1)+(10^(n-2)-1)+...+(10-1)... + 81n
ta có 10^k - 1 = (10-1)[10^(k-1)+...+10+1] chia hết cho 9 =>9[(10^(n-1)-1) +(10^(n-2)-1) +... +(10-1) +(1-1)] chia hết cho 81 =>9[(10^(n-1)-1)+(10^(n-2)-1)+...+(10-1)... + 81n chia hết cho 81 =>đpcm.
21 tháng 7 2015
Ta có :
Cho biểu thức tính trên là A
A = 10n + 72n - 1 = 10n - 1 + 72n
10n - 1 = 99...9 (có n-1 chữ số 9) = 9x(11..1) (có n chữ số 1)
A = 10n - 1 + 72n = 9x(11...1) + 72n => A : 9 = 11..1 + 8n = 11...1 -n + 9n
Ta thấy: 11...1 có n chữ số 1 có tổng các chữ số là n
=> 11..1 - n chia hết cho 9
=> A : 9 = 11..1 - n + 9n chia hết cho 9
Vậy A chia hết cho 81
20 tháng 1 2016
nó cũng dễ thật nhưng mà bạn bich duong thien ty cũng giỏi thật !
NT
1
21 tháng 11 2015
ta có :
cho biểu thức tính trên là A
A=10n+72n-1=10n-1+72n
10n-1=9999...99(có n-1 cs 9) =9.(111..11)( có n chữ số 1)
A=10n-1+72n=9.(111...1)+72n
=>A:9=111...11-n+9n
ta thấy : 11..11 coa n chữ số 1 có tổng các chữ số là n
=>11..1-n chia hết cho 9
=>A:9=11..1-n+9n chia hết cho 9
vậy A chia hết cho 81
TT
1
24 tháng 11 2015
10^n+72n-1
=10^n-1+72n
=(10-1)[10^(n-1)+10^(n-2)+...+10+1]+72n
=9[10^(n-1)+10^(n-2)+...+10+1]-9n+81n
=9[10^(n-1)+10^(n-2)+...+10+1-n]+81n
=9[(10^(n-1)-1)+(10^(n-2)-1)+...+(10-1)... + 81n
Ta có:
10^k - 1 = (10-1)[10^(k-1)+...+10+1] chia hết cho 9
=>9[(10^(n-1)-1) +(10^(n-2)-1) +... +(10-1) +(1-1)] chia hết cho 81
=>9[(10^(n-1)-1)+(10^(n-2)-1)+...+(10-1)... + 81n chia hết cho 81
=>đpcm.
NT
4
S
20 tháng 11 2018
10n+72-1=10n-1-9n+81n
=999.....99(n chữ số)-9n+81n
=9(1111...1(n chữ số)+n)+81n
Ta dễ thấy rằng 111..1(n chữ số) và n có cùng số dư khi chia cho 9
nên 1111...1(n chữ số)-n chia hết cho 9
=> 9(111...1(n chữ số)-n) chia hết cho 81
Mà 81n cũng chia hết cho 81
=> 10n+72n-1 chia hết cho 81 với
n E N
NN
0
HM
3
T
13 tháng 12 2016
A = 10ⁿ + 72n - 1 = 10ⁿ - 1 + 72n
10ⁿ - 1 = 99...9 (có n-1 chữ số 9) = 9x(11..1) (có n chữ số 1)
A = 10ⁿ - 1 + 72n = 9x(11...1) + 72n => A : 9 = 11..1 + 8n = 11...1 -n + 9n
thấy 11...1 có n chữ số 1 có tổng các chữ số là n => 11..1 - n chia hết cho 9
=> A : 9 = 11..1 - n + 9n chia hết cho 9
=> A chia hết cho 81
28 tháng 12 2018
A = 10ⁿ + 72n - 1 = 10ⁿ - 1 + 72n
10ⁿ - 1 = 99...9 (có n-1 chữ số 9) = 9x(11..1) (có n chữ số 1)
A = 10ⁿ - 1 + 72n = 9x(11...1) + 72n => A : 9 = 11..1 + 8n = 11...1 -n + 9n
thấy 11...1 có n chữ số 1 có tổng các chữ số là n => 11..1 - n chia hết cho 9
=> A : 9 = 11..1 - n + 9n chia hết cho 9
=> A chia hết cho 81
6 tháng 1 2017
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10^n+72n-1
=10^n-1+72n
=(10-1)[10^(n-1)+10^(n-2)+...+10+1]+72n
=9[10^(n-1)+10^(n-2)+...+10+1]-9n+81n
=9[10^(n-1)+10^(n-2)+...+10+1-n]+81n
=9[(10^(n-1)-1)+(10^(n-2)-1)+...+(10-1)... + 81n
ta có 10^k - 1 = (10-1)[10^(k-1)+...+10+1] chia hết cho 9 =>9[(10^(n-1)-1) +(10^(n-2)-1) +... +(10-1) +(1-1)] chia hết cho 81 =>9[(10^(n-1)-1)+(10^(n-2)-1)+...+(10-1)... + 81n chia hết cho 81 =>đpcm