a,  A = 1 + 5 3 + 5 5 + 5 7 + . . . + 5 99

B = 5 4 + 5 6 + 5 8 + . . . + 5 100 =  5 . ( 5 3 + 5 5 + 5 7 + . . . + 5 99 ) = 5(A – 1)

A + B – 1 =  5 3 + 5 4 + . . . + 5 100

5(A + B – 1) =  5 4 + 5 5 + . . . + 5 100 + 5 101

4(A + B – 1) = 5(A + B – 1) – (A + B – 1) =  5 101 - 5 3

=> A + B – 1 =  5 101 - 5 3 4

=> A + 5(A – 1) –1 =  5 101 - 5 3 4 => 6A – 6 =  5 101 - 5 3 4

=> A – 1 =  5 101 - 5 3 24

=> A =  5 101 - 5 3 + 24 24

b,  A = 1 - 2 + 2 2 - . . . - 2 2007

A = 1 + 2 2 + . . . + 2 2006 - 2 + 2 3 + . . . + 2 2007

A = ( 1 + 2 2 + . . . + 2 2006 ) - 2 . 1 + 2 2 + . . . + 2 2006

A = - 1 + 2 2 + . . . + 2 2006

Đặt  B = - 2 + 2 3 + . . . + 2 2007 =  - 2 . 1 + 2 2 + . . . + 2 2006 = 2A

A + B =  - 1 + 2 + 2 2 + . . . + 2 2006 + 2 2007

2(A+B) =  - 2 + 2 2 + . . . + 2 2006 + 2 2007 + 2 2008

A+B = 2(A+B)–(A+B) =  - 2 2008 - 1

=> A+2A =  - 2 2008 - 1

=> 3A =  - 2 2008 - 1

=> A =  - ( 2 2008 - 1 ) 3

c,  A = 7 + 7 3 + 7 5 + 7 7 + . . . + 7 1999

Đặt B =  7 2 + 7 4 + 7 6 + . . . + 7 1999 + 7 2000 =  7 ( 7 + 7 3 + 7 5 + 7 7 + . . . + 7 1999 ) = 7A

A+B =  7 + 7 2 + 7 3 + . . . + 7 1999 + 7 2000

7(A+B) =  7 2 + 7 3 + . . . + 7 1999 + 7 2000 + 7 2001

7(A+B) – (A+B) =  ( 7 2 + 7 3 + . . . + 7 1999 + 7 2000 + 7 2001 )  –  ( 7 + 7 2 + 7 3 + . . . + 7 1999 + 7 2000 )

6(A+B) =  7 2001 - 7

A+B =  7 2001 - 7 6

=> A + 7A =  7 2001 - 7 6 => 8A =  7 2001 - 7 6 => A =  7 2001 - 7 48

Ta xét biểu thức \(A_1=7+7^2+7^3\) \(=7\left(1+7+7^2\right)\) \(=57.7⋮57\)

\(A_2=7^4+7^5+7^6\) \(=7^4\left(1+7+7^2\right)\) \(=57.7^4⋮57\)

...

\(A_{40}=7^{118}+7^{119}+7^{120}\) \(=7^{118}\left(1+7+7^2\right)⋮57\)

Vậy \(A=\sum\limits^{40}_{i=1}A_i\) đương nhiên chia hết cho 57 (đpcm)

bài kt cuối kì phải tự làm  bạn ơi

\(A=7+7^2+7^3+...+7^{120}\)

  \(=\left(7+7^2+7^3\right)+\left(7^4+7^5+7^6\right)+...+\left(7^{118}+7^{119}+7^{120}\right)\)

  \(=7.\left(1+7+7^2\right)+7^4.\left(1+7+7^2\right)+...+7^{118}.\left(1+7+7^2\right)\)

  \(=7.57+7^4.57+..+7^{118}.57\)

   \(=57.\left(7+7^4+...+7^{118}\right)\)

⇒ A chia hết cho 57

Bài 1:

\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)

\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)

Bài 2:

\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)

đúng rùi

b) Để 4x + 19 chia hết cho x + 1 thì 15 chia hết cho x + 1

--> x + 1 là ước của 15

TH1: x + 1 = 15 <=> x = 14

TH2: x + 1 = 1 <=> x = 0

TH3: x + 1 = 3 <=> x = 2

TH4: x + 1 = 5 <=> x= 4

7⁶+7⁵-7⁴

=7⁴.7²+7⁴.7-7⁴

=7⁴.(7²+7-1)=7⁴.55 chia hết cho 55

A=1+5+5²+......+5⁵⁰

=>5A=5+5²+5³+......+5⁵¹

=>5A-A=(5+5²+5³+.....+5⁵¹)-(1+5+5²+.....+5⁵⁰)

=>4A=5⁵¹-1

=>A=(5⁵¹-1)/4

b) 5A=5+5^2+5^3+...+5^50+5^51

5A-A=4A=5^51-1

Suy ra A=5^51-1/4

???

Lời giải:

\(A=\frac{1}{2}+\frac{1}{33}+\frac{1}{34}+\frac{1}{35}+\frac{1}{51}+\frac{1}{53}+\frac{1}{55}+\frac{1}{57}+\frac{1}{59}\)

Ta có:

\(\frac{1}{33}+\frac{1}{34}+\frac{1}{35}< \frac{1}{30}+\frac{1}{30}+\frac{1}{30}=\frac{3}{30}=\frac{1}{10}\)

\(\frac{1}{51}+\frac{1}{53}+\frac{1}{55}+\frac{1}{57}+\frac{1}{59}< \frac{1}{50}+\frac{1}{50}+\frac{1}{50}+\frac{1}{50}+\frac{1}{50}=\frac{5}{50}=\frac{1}{10}\)

Cộng theo vế:

\(\frac{1}{33}+\frac{1}{34}+\frac{1}{35}+\frac{1}{51}+\frac{1}{53}+\frac{1}{55}+\frac{1}{57}+\frac{1}{59}< \frac{2}{10}=\frac{1}{5}\)

Suy ra \(A< \frac{1}{2}+\frac{1}{5}=\frac{7}{10}\)

Ta có đpcm.