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lam phamnguyenhoang 15/08/2017 at 21:2760 thank
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Ngu Ngu Ngu 15/04/2017 at 08:04Put \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)
We see:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right).n}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Rightarrow A< 1-\dfrac{1}{n}< 1\)
Conclude:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\) (The thing must prove)
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Nguyễn Hưng Phát 03/12/2017 at 11:23
We have:9*9*9*........*9 (345 number 9)
=9345=9344 * 9=(92)172 * 9=81172 * 9=..............1 * 9=...............9
Answer:the last number of 9*9*9*...........*9 is:9
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Help you solve math 12/08/2017 at 15:2475+15+25+85+60+50+40+50+10
=(75+25)+(15+85)+(60+40)+(50+50)+10
=100+100+100+100+10
=410
This is a quick caclulation
Wish you good study
phanhuytien selected this answer. -
math you like 13/08/2017 at 08:43=410 nhé
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Vương Tuấn Khải _ Tiểu Bàng Giải 14/08/2017 at 12:36=75 + 15 + 25 + 85 + 60 + 50 + 40 + 50 + 10
= 75 + 15 + 25 + 85 + 60 + 40 + 50 + 50 + 10
= 100 + 100 + 100 + 10
= 400 + 10
= 410
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Dao Trong Luan 21/08/2017 at 15:58= 862.38 + 862.62
= 862.\(\left(38+62\right)\)
= 862.100
= 86200
Selected by MathYouLike -
Nguyễn Huy Hoàng 21/08/2017 at 15:59Help you solve math is that right
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Phương Thuý 2k7 22/10/2019 at 10:57
Kết bn kiểu j zậy
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Dương Ngọc Thiện 08/04/2020 at 14:25
ơ hay kì quá
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băng 28/01/2020 at 15:17
kb luôn
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There are : (200 - 1) : 1 + 1 = 200 numbers from 1 to 200
We have a sequence of two-digit numbers are as follows :
2 ; 12 ; 22 ; 32 ; ....... ; 192
Apply a formula to count numbers , we have :
(192 - 2) : 10 + 1 = 20
There are also numbers with 2 digits such as
20 ; 21 ; 23 ; 24 ; .... ; 29 ; 120 ; 121 ; 122 ; 123 ; ...... 129
And there are have 18 also numbers .
So , from these information , we have 38 numbers have digit 2
We have :
200 - 38 = 162 (numbers)
There are 162 numbers from 1 to 200 do not have the digit 2
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Kokone 04/04/2017 at 14:18There are : (200 - 1) : 1 + 1 = 200 numbers from 1 to 200
We have a sequence of two-digit numbers are as follows :
2 ; 12 ; 22 ; 32 ; ....... ; 192
Apply a formula to count numbers , we have :
(192 - 2) : 10 + 1 = 20
There are also numbers with 2 digits such as
20 ; 21 ; 23 ; 24 ; .... ; 29 ; 120 ; 121 ; 122 ; 123 ; ...... 129
And there are have 18 also numbers .
So , from these information , we have 38 numbers have digit 2
We have :
200 - 38 = 162 (numbers)
There are 162 numbers from 1 to 200 do not have the digit 2
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Nguyen Van Hung 27/06/2017 at 20:34
There are : (200 - 1) : 1 + 1 = 200 numbers from 1 to 200
We have a sequence of two-digit numbers are as follows :
2 ; 12 ; 22 ; 32 ; ....... ; 192
Apply a formula to count numbers , we have :
(192 - 2) : 10 + 1 = 20
There are also numbers with 2 digits such as
20 ; 21 ; 23 ; 24 ; .... ; 29 ; 120 ; 121 ; 122 ; 123 ; ...... 129
And there are have 18 also numbers .
So , from these information , we have 38 numbers have digit 2
We have :
200 - 38 = 162 (numbers)
There are 162 numbers from 1 to 200 do not have the digit 2 .
