1.a)A = (1 - 1/3)(1-2/5)...(1-5/5)....(1-9/5)

      =(1-1/3)....0.....(1-9/5)

      =0

     =>đpcm.

b)ta xét:

1/2² = 1/2x2 < 1/1x2

.............

1/8² = 1/8x8 <1/7x8

=>B < 1/1x2 + 1/2x3 ... + 1 + 1/7x8

<=> B <1 - 1/2 + 1/2  - 1/3  + ... + 1/7 - 1/8

<=> B < 1 - 1/8 = 7/8 < 1

=> B < 1 => đpcm

2.a) Đặt m = 2007(2006+2007) = 2006(2006 + 2007) + (2006+2007)

      Đặt n = 2006(2007+2008) = 2006(2006+2007) + (2006 + 2006)

Ta thấy : (2006+2007) > (2006 + 2006) => m > n , áp dụng công thức "a.d > c.d <=> a/b > b/d (a,c thuộc Z// b,d thuộc N)

=> A > B

   b)ta có: D = 196 + 197/197 + 198 = (196/197+198) + (197/197+198) < 196/197 + 197/198 = C

=> C > D

c)gọi 20¹⁰ là a

ta thấy : (a + 1)(a-3) = (a - 1)(a - 3) + 2(a - 3) < (a - 1)(a - 3) + 2(a - 1) = (a - 1)(a - 1)

áp dụng: ad > bc <=> a/b > c/d ( a,b,c,d thuộc Z// b,d > 0)

=> E > F

`A = 2 + 2^2+ ... + 2^2017`

`=> 2A = 2^2 + 2^3 + ... + 2^2018`

`=> 2A - A = (2^2 + 2^3 + ... + 2^2018) - (2 + 2^2 + ... +2^2017)`

`=> A         = 2^2018 - 2`

`B = 1 + 3^2 + ... + 3^2018`

`=> 3^2B = 3^2 + 3^4 + ... + 3^2020`

`=> 9B-B =(3^2 + 3^4 + ... + 3^2020) - (1 + 3^2 + ... + 3^2018`

`=> 8B     = 3^2020 - 1`

`=> B       = (3^2020 - 1)/8`

`C = 5 + 5^2 - 5^3 + ... + 5^2018`

`=> 5C = 5^2 + 5^3 - 5^4 + ... +5^2019`

`=> 5C + C = ( 5^2 + 5^3 - 5^4 + ... 5^2019) + (5 + 5^2 - 5^3 + ... + 5^2018)`

`=> 6C = 55 + 5^2019`

`=> C  = (5^2019 + 55)/6`

Mình làm ngắn gọn nhé.

\(A=1+2+2^2+...+2^{50}\)

\(\Rightarrow2A=2+2^2+...+2^{51}\)

\(\Rightarrow2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}\)

\(\Rightarrow A=2^{51}-1\)

\(B=1+3+...+3^{66}\)

\(3B=3+3^2+...+3^{67}\)

\(2B=3+3^2+...+3^{67}-1-3-...-3^{66}\)

\(2B=3^{67}-1\)

\(B=\frac{3^{67}-1}{2}\)

\(4\cdot5^2-3^2\cdot\left(2021^0+3^2\right)\)
\(=4\cdot25-9\cdot\left(1+9\right)\)
\(=100-9\cdot10\)
\(=100-90\)
\(=10\)

4. 5²- 3². ( 2021⁰+ 3²)

= 4 . 25 - 9 . ( 1 + 9 )

= 100 - 9 . 10

= 100-90

= 10

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

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

\(=2+2^2\left(1+2+2^2\right)+...+2^{98}\left(1+2+2^2\right)\)

\(=2+7\cdot\left(2^2+2^5+...+2^{98}\right)\)

=>A không chia hết cho 7 mà là chia 7 dư 2 nha bạn

`A=1+2^2 +2^3 +...+2^10`

`2A=2+2^3 +2^4 +...+2^11`

`A=2+2^3 +2^4 +...+2^11 -1-2^2 -2^3 -...-2^10`

`A=2+2^11 -1-2^2`

`A=2+2048-1-4`

`A=2045`

Đặt: \(A=1+2^2+2^3+...+2^{10}\)

\(\Rightarrow2A=2\cdot\left(1+2^2+2^3+...+2^{10}\right)\)

\(\Rightarrow2A=2+2^3+2^4+...+2^{11}\)

\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)

\(\Rightarrow A=2+2^3+2^4+...+2^{11}-1-2^2-2^3-...-2^{10}\)

\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2^{10}-2^{10}\right)+\left(2+2^{11}-1-2^2\right)\)

\(\Rightarrow A=0+0+0+...+2+2^{11}-1-2^2\)

\(\Rightarrow A=2+2^{11}-1-4\)

\(\Rightarrow A=2^{11}-3\)

Ta có công thức tổng quát như sau:

\(A=n^k+n^{k+1}+n^{k+2}+...+n^{k+x}\Rightarrow A=\dfrac{n^{k+x+1}-n^k}{n-1}\)

Áp dụng ta có:

\(A=1+4+4^2+...+4^6=\dfrac{4^7-1}{3}\) 

\(\Rightarrow B-3A=4^7-3\cdot\dfrac{4^7-1}{3}=1\)

______

\(A=2^0+2^1+...+2^{2008}=2^{2009}-1\)

\(\Rightarrow B-A=2^{2009}-2^{2009}+1=1\)

_____

\(A=1+3+3^2+....+3^{2006}=\dfrac{3^{2007}-1}{2}\)

\(\Rightarrow B-2A=3^{2007}-2\cdot\dfrac{3^{2007}-1}{2}=1\)