S=1/2⁰+(1/2¹+1/2^2-1)+(1/2²+...+1/2^3-1)+...+(1/2⁹⁹+...+1/2^100-1)         (100 cặp)

S<1/2⁰.2⁰+1/2¹.2¹+1/2².2²+...+1/2⁹⁹.2⁹⁹

S<1+1+1+...+1 (100 số 1)

S<100.1

S<100 (ĐPCM)

Ta có:

\(2\left(\sqrt{n+1}-\sqrt{n}\right)=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{2}{\sqrt{n+1}+\sqrt{n}}< \dfrac{2}{2\sqrt{n}}=\dfrac{1}{\sqrt{n}}\)

\(\Rightarrow S>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{101}-\sqrt{100}\right)=2\left(\sqrt{101}-1\right)>18\)

\(2\left(\sqrt{n}-\sqrt{n-1}\right)=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)\left(\sqrt{n}+\sqrt{n-1}\right)}{\left(\sqrt{n}+\sqrt{n-1}\right)}=\dfrac{2}{\sqrt{n}+\sqrt{n-1}}>\dfrac{2}{2\sqrt{n}}=\dfrac{1}{\sqrt{n}}\)

\(\Rightarrow S< 1+2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)=1+2\left(\sqrt{100}-1\right)=19\)

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`

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S = 2^1 + 2^2 + 2^3 +...+ 2^100

S = (2^1 + 2^2 +2^3 + 2^4) + ... + (2^97 + 2^98 + 2^99 + 2^100)

S = 2(1 + 2 + 2^2 + 2^3 ) + ...+ 2^97( 1 + 2 + 2^2 + 2^3)

S = 2x15 +...+ 2^97x15

S = 15( 2...2^97) chia hết cho 15

Do 15 chia hết cho 3 mà S chia hết cho 15

=> S chia hết cho 3 

Vậy s chia hết cho 3 và 15