a) S = 2 + 2² + 2³ + 2⁴ +.....+ 2⁹ + 2¹⁰

   = (2 + 2²) + (2³ + 2⁴) +.....+ (2⁹ + 2¹⁰)

   = 2(1 + 2) + 2³(1 + 2) +....+ 2⁹(1 + 2)

   = 3.(2 + 2³ +.... + 2⁹) chia hết cho 3

   => S = 2 + 2² + 2³ + 2⁴ +.....+ 2⁹ + 2¹⁰ chia hết cho 3 (Đpcm)

b) 1+3²+3³+3⁴+...+3⁹⁹

=(1+3+3²+3³)+....+(3⁹⁶+3⁹⁷+3⁹⁸+3⁹⁹)

=40+.........+3⁹⁶.40

=40.(1+....+3⁹⁶) chia hết cho 40 (đpcm)

ai trả lời giúp mình mình k cho

BÀI 1:

S = 2 + 2² + 2³ + 2⁴ + ..... + 2¹⁰

= (2 + 2²) + ( 2³ + 2⁴) + ..... + (2⁷ + 2⁸) + (2⁹ + 2¹⁰)

= 2(1 + 2) + 2³(1 + 2) + ..... + 2⁷(1 + 2) + 2⁹(1 + 2)

= 3(2 + 2³ + .... + 2⁷ + 2⁹)    \(⋮3\)

BÀI 2:

1 + 3 + 3² + 3³ + ....... + 3⁹⁹

= (1 + 3 + 3² + 3³) + ..... + (3⁹⁶ + 3⁹⁷ + 3⁹⁸ + 3⁹⁹)

= (1 + 3 + 3² + 3³) + ..... + 3⁹⁶(1 + 3 + 3² + 3³)

= 40(1 + 3⁴ + ..... + 3⁹⁶)     \(⋮40\)

\(\frac{M}{3}=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)

\(\frac{2M}{3}=M-\frac{M}{3}=\frac{1}{3}-\frac{1}{3^{100}}\)

\(2M=1-\frac{1}{3^{99}}\Rightarrow M=\frac{1}{2}-\frac{1}{2.3^{99}}

\(M=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right).2.3.4...2018\)

\(\Rightarrow M=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right).2.3.4...673.674...2018\)

Vì \(\hept{\begin{cases}M⋮3\\M⋮673\end{cases}}\) mà \(\left(3,673\right)=1\) nên \(M⋮2019\left(đpcm\right)\)

\(M=\left[\left(1+\frac{1}{2018}\right)+\left(\frac{1}{2}+\frac{1}{2017}\right)+...+\left(\frac{1}{1008}+\frac{1}{1011}\right)+\left(\frac{1}{1009}+\frac{1}{1010}\right)\right].\)\(2.3...1008.1009.1010.1011...2017.2018\)

\(=\left(\frac{2019}{2018}+\frac{2019}{2.2017}+...+\frac{2019}{1008.1011}+\frac{2019}{1009.1010}\right).2.3...1008.1009.1010.1011...2017.2018\)

\(=2019\left(\frac{1}{2018}+\frac{1}{2.2017}+...+\frac{1}{1008.1011}+\frac{1}{1009.1010}\right).2...1008.1009.1010.1011...2017.2018\)

\(=2019.\left(2...2017+3...2016.2018+...+2.3...1007.1009.1011...2018+2.3....1008.1011...2018\right)\)

Chia hết cho 2019

❖︵Ňɠυүễη Çɦâυ Ƭυấη Ƙїệт♔:Nể phục nể phục!Đã biết  \(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2018}\) là số nguyên chưa ạ

\(10M=1+\frac{1}{10}+\frac{1}{10^2}+...+\frac{1}{10^{2008}}\)

\(9M=10M-M=1-\frac{1}{10^{2009}}\Rightarrow M=\frac{1}{9}-\frac{1}{9.10^{2009}}< \frac{1}{9}\)

Ta có:3.A=1+1/3+1/3^2+...+1/3^97 +1/3^98

=>3.A - A=(1+1/3+1/3^2+...+1/3^98 + 1/3^99)-(1/3+1/3^2 +1/3^3+...+1/3^98+1/3^99)

=>2.A=1-1/3^99

=>A=1/2 -1/3^99.1/2 <1/2

Vậy ... T I C K cho mình với nha

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