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Ta có: \(S=\frac{1}{5^2}+\frac{2}{5^3}+\cdots+\frac{99}{5^{100}}\)
=>\(5S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}\)
=>\(5S-S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}-\frac{1}{5^2}-\frac{2}{5^3}-\cdots-\frac{99}{5^{100}}=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
=>\(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
Ta có: \(A=\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}\)
=>\(5A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}\)
=>\(5A-A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}-\frac{1}{5^2}-\frac{1}{5^3}-\cdots-\frac{1}{5^{99}}\)
=>\(4A=\frac15-\frac{1}{5^{99}}=\frac{5^{98}-1}{5^{99}}\)
=>\(A=\frac{5^{98}-1}{4\cdot5^{99}}\)
Ta có: \(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
\(=\frac15+\frac{5^{98}-1}{4\cdot5^{99}}-\frac{99}{5^{100}}=\frac15+\frac{5^{99}-5-396}{4\cdot5^{100}}=\frac15+\frac{1}{4\cdot5}-\frac{401}{4\cdot5^{100}}\)
=>\(4S<\frac15+\frac{1}{20}=\frac{4}{20}+\frac{1}{20}=\frac{5}{20}=\frac14\)
hay S<1/16
Ta có: \(S=\frac{1}{5^2}+\frac{2}{5^3}+\cdots+\frac{99}{5^{100}}\)
=>\(5S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}\)
=>\(5S-S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}-\frac{1}{5^2}-\frac{2}{5^3}-\cdots-\frac{99}{5^{100}}=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
=>\(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
Ta có: \(A=\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}\)
=>\(5A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}\)
=>\(5A-A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}-\frac{1}{5^2}-\frac{1}{5^3}-\cdots-\frac{1}{5^{99}}\)
=>\(4A=\frac15-\frac{1}{5^{99}}=\frac{5^{98}-1}{5^{99}}\)
=>\(A=\frac{5^{98}-1}{4\cdot5^{99}}\)
Ta có: \(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
\(=\frac15+\frac{5^{98}-1}{4\cdot5^{99}}-\frac{99}{5^{100}}=\frac15+\frac{5^{99}-5-396}{4\cdot5^{100}}=\frac15+\frac{1}{4\cdot5}-\frac{401}{4\cdot5^{100}}\)
=>\(4S<\frac15+\frac{1}{20}=\frac{4}{20}+\frac{1}{20}=\frac{5}{20}=\frac14\)
hay S<1/16
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!!!!!
Lời giải:
$S=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}$
$5S=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+....+\frac{99}{5^{99}}$
$5S-S=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}$
$4S+\frac{99}{5^{100}}=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}$
$5(4S+\frac{99}{5^{100}})=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}$
$5(4S+\frac{99}{5^{100}})-(4S+\frac{99}{5^{100}})=1-\frac{1}{5^{99}}$
$4(4S+\frac{99}{5^{100}})=1-\frac{1}{5^{99}}$
$16S=1-\frac{1}{5^{99}}-\frac{99.4}{5^{100}}<1$
$\Rightarrow S< \frac{1}{16}$