*)S=2+2²+2³+2⁴+.....+2⁸

Vì các số hạng của S chia hết chia hết cho 2

*) S=2+2²+2³+2⁴+.....+2⁸

=> S=(2+2²)+(2³+2⁴)+.....+(2⁷+2⁸)

=> S=2(1+2)+2³(1+2)+....+2⁷(1+2)

=> S=2.3+2³.3+.....+2⁷.3

=> S=3(2+2³+....+2⁷)

=> S chia hết cho 3

Ta có 2 và 3 là 2 số nguyên tố cùng nhau => S chia hết cho 2.3=6

=> S chia hết cho -6 (đpcm)

\(S=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8\)

 \(=2\left(1+2\right)+2^3\left(1+2\right)+2^5\left(1+2\right)+2^7\left(1+2\right)\)

\(=2.3+2^3.3+2^5.6+2^7.3\)

\(=6+2^2.6+2^4.6+2^6.6⋮6\)

Vậy \(S⋮6\)

\(#hoktot\)

Dễ dàng cmđ S chia hết -2 (1)

Ta đi cm S chia hết 3

Có 2+2^2=(2+1)+(2^2-1)

Có 2+1 chia hết cho 3

2^2-1 chia hết 2+1=3 ( Do 2 chẵn )

Từ 2 điều trên => 2+2^2 chia hết 3

Tương tự 2^3+2^4 ; 2^5+2^6;2^7+2^8 chia hết 3

=> S chia hết 3 (2)

(1);(2) => S chia hết -6 (vì UCLN(3;-2)=1)

Vậy...

Chúc học tốt nhaaa

B = (1 + 3) + (3²+3³)+.....+(3⁸⁹+3⁹⁰)

  = 4 + 3² .(1 + 3) + .....+3⁹⁰.(1+3)

 = 1 .4 + 3².4 + ..... +3⁹⁰.4

= 4.(1 + 3² + .... +3⁹⁰) chia hết cho 4

\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)

\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)

\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)

\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)

\(S=2^0+2^2+...+2^{2014}.\)

\(S=\left(2^0+2^2+2^4+2^6\right)+.....+\left(2^{2008}+2^{2010}+2^{2012}+2^{2014}\right)\)

\(S=17+.....+2^{2008}.17\)

\(S=17.\left(2^0+...+2^{2008}\right)\)

\(\Leftrightarrow S⋮17\left(đpcm\right)\)

\(S=2^0+2^2+...+2^{2014}.\)

\(S=\left(2^0+2^2+2^4\right)+....+\left(2^{2010}+2^{2012}+2^{2014}\right)\)

\(S=21+....+2^{2010}.21\)

\(S=21.\left(2^0+...+2^{2010}\right)\)

\(S=7.3.\left(2^0+....+2^{2010}\right)\)

\(\Leftrightarrow S⋮7\left(đpcm\right)\)

S = 2⁰ + 2² + 2⁴ + 2⁶ + 2⁸ + ... + 2²⁰¹⁴

S = (2⁰ + 2² + 2⁴) + (2⁶ + 2⁸ + 2¹⁰) + ... + (2²⁰¹⁰ + 2²⁰¹² + 2²⁰¹⁴)

S = (2⁰ + 2² + 2⁴) + 2⁶(2⁰ + 2² + 2⁴) + ... + 2²⁰¹⁰(2⁰ + 2² + 2⁴)

S = (2⁰ + 2² + 2⁴)(2⁶ + ... + 2²⁰¹⁰) 

S =         21   .  (2⁶ + ... + 2²⁰¹⁰) 

Vì 21 \(⋮\)7 nên 21 . (2⁶ + ... + 2²⁰¹⁰)  \(⋮\)7 => S \(⋮7\)

S=2⁰+2²+...+2²⁰¹⁴.

S=(2⁰+2²+2⁴+2⁶)+.....+(2²⁰⁰⁸+2²⁰¹⁰+2²⁰¹²+2²⁰¹⁴)

S=17+.....+2²⁰⁰⁸.17

S=17.(2⁰+...+2²⁰⁰⁸)

=> S \(⋮\)17

Đặt A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

A=\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)

Ta thấy :

\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\)

\(\dfrac{1}{100.100}< \dfrac{1}{99.100}\)

\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

Nhân xét :

\(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};\)

\(...;\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}+...+\)

\(\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow A< 1-\dfrac{1}{100}\)

\(\Rightarrow A< \dfrac{99}{100}\)

Vì \(A< \dfrac{99}{100}< 1\)

\(\Rightarrow A< 1\)

Bài 1)

Đặt \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\)
Ta thấy:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4};....;\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\)\(\Rightarrow\) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\) < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
\(\Rightarrow\) A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow\) A < \(1-\dfrac{1}{100}\) < 1 \(\Rightarrow\) A < 1

Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\)< 1

Bài 1:

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1\left(đpcm\right)\)

Bài 2: bạn xem lại xem có phải \(S=3^0+3+3^2+...+3^{2002}\) không nhé!

Bài 3:

\(A=5+5^2+..+5^{12}\)

\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)

\(5A=5^2+5^3+...+5^{13}\)

\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)

\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)

\(4A=5^{13}-5\)

\(A=\dfrac{5^{13}-5}{4}\)

a, S = 1+3+3²+3³+...+3³⁹

3S = 3+3²+3³+3⁴+...+3⁴⁰

2S = 3S - S = 3⁴⁰ - 1

=> S = \(\frac{3^{40}-1}{2}\)

b, S = 1+3+3²+3³+...+3³⁹

S = (1+3+3²+3³)+(3⁴+3⁵+3⁶+3⁷)+....+(3³⁶+3³⁷+3³⁸+3³⁹)

S = 1(1+3+3²+3³) + 3⁴(1+3+3²+3³) +.....+ 3³⁶.(1+3+3²+3³)

S = 1.40 + 3⁴.40 +.....+ 3³⁶.40

S = 40.(1+3⁴+...+3³⁶) chia hết cho 40 

S = 4.10.(1+3⁴+...+3³⁶) chia hết cho 4

=> Đpcm

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^7+2^8\right)\)

\(=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^7.\left(1+2\right)\)

\(=3.\left(2+2^3+...+2^7\right)=3.2.\left(1+2^2+2^3+...+2^6\right)\)

\(=6.\left(1+2^2+2^3+...+2^6\right)⋮-6\)

\(S=2+2^2+2^3+...+2^8\)

\(=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^6\left(2+2^2\right)\)

\(=1\cdot6+2^2\cdot6+...+2^6\cdot6\)

\(=6\left(1+2^2+...+2^6\right)=-6\cdot\left(-1\right)\left(1+2^2+...+2^6\right)⋮\left(-6\right)\)