Ta có

\(S=\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}\right)\)

\(S>\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{1}{2}.4=2\)

Vậy S>2

\(S=3+3^2+3^3+3^4+3^5+.....+3^{99}+3^{100}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+......+\left(3^{99}+3^{100}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+.......+3^{99}\left(1+3\right)\)

\(=\left(1+3\right)\left(3+3^3+....+3^{99}\right)\)

\(=4\left(3+3^3+.....+3^{99}\right)\)chia hết cho ( đpcm )

\(s=\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(s=3\left(1+3+3^2+3^3\right)+...+3^{97}\left(1+3+3^2+3^3\right)\)

\(s=\left(1+3+3^2+3^3\right).\left(3+...+3^{97}\right)\)

\(s=120.\left(3+...+3^{97}\right)\)

\(\Rightarrow\)s chia hết cho 120

S = 3⁰ + 3² + 3⁴ + .... + 3²⁰⁰²

Nhân cả hai vế của S với 3² ta được :

3²S = 3² ( 3⁰ + 3² + 3⁴ + .... + 3²⁰⁰² )

= 3² + 3⁴ + 3⁶ + ..... + 3²⁰⁰⁴

Trừ cả hai vế của 3²S cho S ta được :

3²S - S = ( 3² + 3⁴ + 3⁶ + ..... + 3²⁰⁰⁴ ) - ( 3⁰ + 3² + 3⁴ + .... + 3²⁰⁰² )

8S = 3²⁰⁰⁴ - 1

\(\Rightarrow S=\frac{3^{2004}-1}{8}\)

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giải thích rõ ra bn 

S=1-3+3²-...+3⁹⁸-3⁹⁹

=(1-3+3²-3³)+...+(3⁹⁶-3⁹⁷+3⁹⁸-3⁹⁹)

=-20+...+3⁹⁶.(-20)

=-20.(1+....+3⁹⁶)

nên S chhia hết cho(-20) (đpcm) 

=>(1-3+3²-3³)+...+(3⁹⁶-3⁹⁷+3⁹⁸-3⁹⁹)

=> -20+...+3⁹⁶.(-20)

=>-20.(1+....+3⁹⁶)

Nên S chhia hết cho(-20) (đpcm) 

tick nhé

\(S=\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2017}}\)

\(4S=1+\frac{1}{4}+...+\frac{1}{4^{2016}}\)

\(4S-S=\left(1+\frac{1}{4^1}+...+\frac{1}{4^{2016}}\right)-\left(\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2017}}\right)\)

\(3S=1-\frac{1}{4^{2017}}< 1\)

\(\Rightarrow S< \frac{1}{3}\left(đpcm\right)\)