![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$S=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}$
$5S=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+....+\frac{99}{5^{99}}$
$5S-S=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}$
$4S+\frac{99}{5^{100}}=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}$
$5(4S+\frac{99}{5^{100}})=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}$
$5(4S+\frac{99}{5^{100}})-(4S+\frac{99}{5^{100}})=1-\frac{1}{5^{99}}$
$4(4S+\frac{99}{5^{100}})=1-\frac{1}{5^{99}}$
$16S=1-\frac{1}{5^{99}}-\frac{99.4}{5^{100}}<1$
$\Rightarrow S< \frac{1}{16}$
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : S = ( 5 + 52 ) + ( 53 + 54 ) + .... + ( 599 + 5100 )
= 5 ( 1 + 5 ) + 53 ( 1 + 5 ) + ..... + 599 ( 1 + 5 )
= 5.6 + 53.6 + .... + 599.6
= 6 ( 5 + 53 + ... + 599 )
Vì 6 chia hết cho 6 nên 6 ( 5 + 53 + ... + 599 ) chia hết cho 6
Hay S chia hết cho 6 ( đpcm )
Ta có A=5+52+53+...+599+5100=(5+52)+(53+54)+...+(599+5100)
A=5.(1+5)+53.(1+5)+599.(1+5)
A=5.6+53.6+...+599.6
A=6.(5+53+...+599) sẽ chia hết cho 6
mik nha bài nay mik làm HSG lớp 6 quen rùi!!!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=5\left(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{100^2}\right)\)Ta có :
\(S< 5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=5\left(1-\frac{1}{100}\right)< 5\)
\(S>5\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\right)=5\left(\frac{1}{2}-\frac{1}{101}\right)>2\)
\(\Rightarrow2< S< 5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\); \(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\); ...; \(\frac{1}{100^2}< \frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)
=> S < \(5\left(1-\frac{1}{100}\right)=5.\frac{99}{100}< 5.1=5\)=> S<5
Lại có: \(\frac{1}{2^2}>\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\); \(\frac{1}{3^2}>\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\); \(\frac{1}{100^2}>\frac{1}{100.101}=\frac{1}{100}-\frac{1}{101}\)
=> \(S>5\left(\frac{1}{2}-\frac{1}{101}\right)=5.\frac{101-2}{2.101}=\frac{5.99}{2.101}~2,45\)=> S>2
Vậy 2 < S < 5 => Đpcm