Question

Cho : S = 3⁰ + 3² + 3⁴ + 3⁶ + ... + 3²⁰⁰²

 a) Tính S 

 b) Chứng minh S chia hết cho 7

Answer 1

S = 3⁰ + 3² + 3⁴ + 3⁶ + ... + 3²⁰⁰²

S = 1 + 3² + 3⁴ + 3⁶ + ... + 3²⁰⁰²

3²S = 3² + 3⁴ + 3⁶ + ... + 3²⁰⁰⁴

3²S-S= (3² + 3⁴ + 3⁶ + ... + 3²⁰⁰⁴)-( 1 + 3² + 3⁴ + 3⁶ + ... + 3²⁰⁰²)

3²S-S= 3²⁰⁰⁴-1

Hay S(3²-1)=3²⁰⁰⁴-1

=> 8S=3²⁰⁰⁴-1

=> S=(3²⁰⁰⁴-1)/8

Answer 2

 S=(3⁰+3²+3⁴)+...+(3¹⁹⁹⁸+3²⁰⁰⁰+3²⁰⁰²)

S= 91+...+3¹⁹⁹⁸(1+3²+3⁴)

S=91+...+3¹⁹⁹⁸.91

S=91(1+3⁶+...+3¹⁹⁹⁸)

S=13.7.(1+3⁶+...+3¹⁹⁹⁸) chia hết cho 7

Answer 3

a)nhân S với 3² ta dc:

9S=3^2+3^4+...+3^2002+3^2004

=>9S-S=(3^2+3^4+...+3^2004)-(3^0+3^4+...+2^2002)

=>8S=3^2004-1

=>S=3^2004-1/8

b) ta có S là số nguyên nên phải chứng minh 3^2004-1 chia hết cho 7

ta có:3^2004-1=(3⁶)^334-1=(3^6-1).M=7.104.M

=>3²⁰⁰⁴ chia hết cho 7. Mặt khác ƯCLN_(7;8)=1 nên S chia hết cho 7

Answer 4

a) S=3^2004-1:8

b)S chia hết cho 7

P/S : ko nói nhìu

 

Answer 5

a)   S=3⁰+3²+3⁴+3⁶+...+3²⁰⁰²

   3²S=3²+3⁴+3⁶+3⁸+...+3²⁰⁰⁴

3²S-S=(3²+3⁴+3⁶+3⁸+...+3^2004)-(3⁰+3²+3⁴+3⁶+...+3²⁰⁰²⁾

      8S=3²⁰⁰⁴-1

      S=(3²⁰⁰⁴-1) : 8

Answer 6

S= 3^0 +3^2 +3^4 +....+ 3^2002

9S= 3^4 +3^6+.......+3^2004

9S-S=3^2004-1

8S=3^2004-1

S=3^2004-1/8

Answer 7

a) S=3+3^2+3^4+3^6+...+3^2002

 3S=3^2+3^4+3^6+...+3^2002+3^2004

3S-S=3^2004-3

2S=3^2004-3

S=3^2004-3/2

b) S=7 . ( 3+3^2+3^4+3^6+...3^2002) chia hết cho 7

Vậy S chia hết cho 7.

Answer 8

cac ban lam dung roi

Answer 9

   co tat ca bao nhieu nhom vay

Answer 10

a) \(S=3^0+3^2+3^4+...+3^{2002}+3^{2004}\)

\(9S=3^4+3^6+...3^{2004}\)

\(9S-S=3^{2004}-1\)

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

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

b)\(S=\left(3^0+3^2+3^4\right)+...+\left(3^{1998}+3^{2000}+3^{2002}\right)\)

\(S=91+...+3^{1998}\left(1+3^2+3^4\right)\)

\(S=91+3^{1998}.91\)

\(S=91\left(1+3^6+...+3^{1988}\right)\)

\(S=13.7.\left(1+3^6+...+3^{1998}\right)⋮7\)

Answer 11

a,S=(3^2004-1):8.

b)S=3^0+3^2+3^4+......+3^2016.

     =3^0+3^2+3^4+.......+3^2010.(3^0+3^2+3^4).

     =1+9+81+.......+3^2010(1+9+81)

     =91+.....+3^2010. 91

     =91.(1+.....+3^2010)

                  Vì 91 chia hết cho 7  nên S chia hết cho 7.

                   Vậy S chia hết cho 7.

Answer 12

Ta có: S = 3⁰ + 3² + 3⁴ + 3⁶ + … + 3²⁰⁰² (1)
Nhân cả hai vế của (1) cho 9, ta được:
9S = 3²(3⁰ + 3² + 3⁴ + 3⁶ + … + 3²⁰⁰²)
9S = 3² + 3⁴ + 3⁶ + 3⁸ + … + 3²⁰⁰⁴ (2)
Lấy (2) - (1), ta được:
9S - S = (3² + 3⁴ + 3⁶ + 3⁸ + … + 3²⁰⁰⁴) - (3⁰ + 3² + 3⁴ + 3⁶ + … + 3²⁰⁰²)
8S = 3²⁰⁰⁴ - 3⁰
8S = 3²⁰⁰⁴ - 1
Khi đó: 
8S - 3²⁰⁰⁴ - 1 = 3²⁰⁰⁴ - 1 - 3²⁰⁰⁴ - 1
8S - 3²⁰⁰⁴ - 1 = -2

Answer 13

k cho mik đã 

Answer 14

a, S = 3^0+3^2+3^4+3^6+...+3^2002

=>9S =3^2+3^4+3^6+3^8+...+3^2002+3^2004

=>9S - S=8S=(3^2+3^4+3^6+3^8+...+3^2002+3^2004)-(3^0+3^2+3^4+3^6+...+3^2002)

=>8S=3^2+3^4+3^6+3^8+...+3^2002+3^2004-3^0-3^2-3^4-3^6-...-3^2002

(bạn rút gọn : vd 3^2 vs -3^20

=>S=3^2004-1/8

b,S=3^0+3^2+3^4+...+3^1998+3^2000+3^2002

S=(3^0+3^2+3^4)+...+3^1998.(3^0+3^2+3^4)

S=91+...+3^1998.91

S=91.(1+...+3^1998)

S=7.13.(1+...+3^1998)

=>S chia hết cho 7(đpcm)

Answer 15

a, S = 3^0+3^2+3^4+...+3^2002

    3^2.S = 3^2+3^4+3^6+...+3^2004

    9S - S = (3^2+3^4+3^6+...+3^2004)-(3^0+3^2+3^4+...+3^2002)

    8S = 3^2+3^4+3^6+...+3^2004-3^0-3^2-3^4+...+3^2002

    S = 3^2004-1/8

b, S = 3^0+3^2+3^4+...+3^1998+3^2000+3^2002

    S = (3^0+3^2+3^4)+...+3^1998.(3^0+3^2+3^4)

    S = 91+...+3^1998.91

    S = 91.(1+...+3^1998) = 7.13.(1+...+3^1998)

=>S chia hết cho 7 (đpcm)