Question

\(S=3^0+3^2+3^4+3^6+.....+3^{2002}\)

Tính S 

Nhanh nha

Answer 1

lớp 1 sao cái này là lớp 6 nha bạn

Answer 2

Nhân với 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 + ... + 3 ^ 2002 )

=> 8S = 3^{2004 - 1} : 8

=> S = 3^{2004- 1} : 8

Answer 3

\(S=1+3^2+3^3+...+3^{2002}\)

\(3S=3+3^2+3^3+3^4+...+3^{2002}+3^{2003}\)

\(3S-S=\left(3+3^2+3^3+3^4+....+3^{2002}+3^{2003}\right)-\left(1+3^2+3^3+...+3^{2002}\right)\)

\(3S-S=3+3^2+3^3+3^4+....+3^{2002}+3^{2003}-1-3^2-3^3-...-3^{2002}\)

\(2S=3^{2003}-1\)

\(S=\frac{3^{2003}-1}{2}\)

Answer 4

    S=1+3²+3⁴+3⁶+...+3²⁰⁰²-

3².S=3²+3⁴+3⁶+3⁸+...+3²⁰⁰⁴

3².S-S= 8S=3²⁰⁰⁴-1

Vậy S=\(\frac{3^{2004}-1}{8}\)

Answer 5

Ta có :

 \(3^2S=3^2+3^4+3^6+....+3^{2004}\)

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

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

Ủng hộ  nha 

Thank you very much

Answer 6

9S = \(9.\left(3^0+3^2+...+3^{2002}\right)\)

=> 9S = \(3^2+3^4+...+3^{2002}+3^{2004}\)

=> 9S - S = 8S = \(\left(3^2+3^4+..+3^{2004}\right)-\left(3^0+3^2+...+3^{2002}\right)\)

=> 8S = \(3^{2004}-1\)

=> S = \(\left(3^{2004}-1\right):8\)

Answer 7

toan lop 1 sao lai co luy thua