Cho S=1/5^2+2/5^3+...+99/5^100.Chứng tỏ rằng S<1/16
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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!!!!!
a) \(S=5+5^2+5^3+5^4+...+5^{99}\)
\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{97}+5^{98}+5^{99}\right)\)
\(=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{97}\left(1+5+5^2\right)\)
\(=5.31+5^4.31+...+5^{97}.31\)
\(=31\left(5+5^4+...+5^{97}\right)⋮31\left(đpcm\right)\)
b) \(S=5+5^2+5^3+5^4+...+5^{99}\)
\(=5+\left(5^2+5^3\right)+\left(5^4+5^5\right)+...+\left(5^{98}+5^{99}\right)\)
\(=5+5\left(5+5^2\right)+5^3\left(5+5^2\right)+...+5^{97}\left(5+5^2\right)\)
\(=5+5.30+5^3.30+...+5^{97}.30\)
\(=5+30.\left(5+5^3+...+5^{97}\right)\)
Mà \(5⋮̸30\) nên \(S⋮̸30\left(đpcm\right)\)
c) Ta có: \(5S=5^2+5^3+5^4+5^5+...+5^{100}\)
\(5S-S=\left(5^2+5^3+5^4+5^5+...+5^{100}\right)-\left(5+5^2+5^3+5^4+...+5^{99}\right)\)
\(4S=5^{100}-5\)
\(\Rightarrow25^x-5=5^{100}-5\)
\(\Rightarrow25^x=5^{100}\)
\(\Rightarrow25^x=25^{50}\)
\(\Rightarrow x=50\)
Lời giải:
$S=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}$
$5S=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+....+\frac{99}{5^{99}}$
$5S-S=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}$
$4S+\frac{99}{5^{100}}=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}$
$5(4S+\frac{99}{5^{100}})=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}$
$5(4S+\frac{99}{5^{100}})-(4S+\frac{99}{5^{100}})=1-\frac{1}{5^{99}}$
$4(4S+\frac{99}{5^{100}})=1-\frac{1}{5^{99}}$
$16S=1-\frac{1}{5^{99}}-\frac{99.4}{5^{100}}<1$
$\Rightarrow S< \frac{1}{16}$