S=1+(-5^1)+(-5^2)+...+(-5^100)
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Ta có :
`5S=5(1/(5^2)+2/(5^3)+3/(5^4)+...+99/(5^100))`
`5S=1/5+2/(5^2)+3/(5^3)+...+99/(5^100)`
`=>5S-S=1/5+2/(5^2)+3/(5^3)+...+99/(5^100)-(1/(5^2)+2/(5^3)+3/(5^4)+...+99/(5^100))`
`4S=1/5+1/(5^2)+1/(5^3)+1/(5^4)+...+1/(5^99) -99/(5^100)`
`20S=5(1/5+1/(5^2)+1/(5^3)+...+1/(5^99)-99/(5^100))`
`20S=1+1/5+1/(5^2)+....+1/(5^98)-99/(5^99)`
`=>20S-4S=(1+1/5+1/(5^2)+...+1/(5^98)-99/(5^99))-(1/5+1/(5^2)+1/(5^3)+...+1/(5^99)-99/(5^100))`
`=>16S=1-99/(5^99)-1/(5^99)-99/(5^100)`
Vì `-99/(5^99)-1/(5^99)-99/(5^100)<0=>1-99/(5^99)-1/(5^99)-99/(5^100)<1`
`=>S<1/16`
S = 1/5 + 1/5² + 1/5³ + ... + 1/5¹⁰⁰
⇒ 5S = 1 + 1/5 + 1/5² + ... + 1/5⁹⁹
⇒ 4S = 5S - S
= (1 + 1/5 + 1/5² + ... + 1/5⁹⁹) - (1/5 + 1/5² + 1/5³ + ... + 1/5¹⁰⁰)
= 1 - 1/5¹⁰⁰
⇒ S = (1 - 1/5¹⁰⁰)/4
S = \(\dfrac{1}{5}\) + \(\dfrac{1}{5^2}\) + \(\dfrac{1}{5^3}\)+...+\(\dfrac{1}{5^{99}}\)+ \(\dfrac{1}{5^{100}}\)
5S = 1 + \(\dfrac{1}{5}\) + \(\dfrac{1}{5^2}\) + \(\dfrac{1}{5^3}\)+...+ \(\dfrac{1}{5^{99}}\)
5S - S = 1 - \(\dfrac{1}{5^{100}}\)
4S = \(\dfrac{5^{100}-1}{4.5^{100}}\)
Lời giải:
$S=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}$
$5S=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+....+\frac{99}{5^{99}}$
$5S-S=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}$
$4S+\frac{99}{5^{100}}=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}$
$5(4S+\frac{99}{5^{100}})=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}$
$5(4S+\frac{99}{5^{100}})-(4S+\frac{99}{5^{100}})=1-\frac{1}{5^{99}}$
$4(4S+\frac{99}{5^{100}})=1-\frac{1}{5^{99}}$
$16S=1-\frac{1}{5^{99}}-\frac{99.4}{5^{100}}<1$
$\Rightarrow S< \frac{1}{16}$