Question

Cho S=3^0+3^2+3^4+3^6+.......+3^2002

A.Tính S

Answer 1

Ta có :

S = 3⁰ + 3² + 3⁴ + 3⁶ + ....... + 3²⁰⁰² ( 1 )

9S = 3² + 3⁴ + 3⁶ + ....... + 3²⁰⁰⁴ ( 2 )

Lấy ( 1 ) - ( 2 ) ta có

9S - S = ( 3² + 3⁴ + 3⁶ + ....... + 3²⁰⁰⁴ ) - ( 3⁰ + 3² + 3⁴ + 3⁶ + ....... + 3²⁰⁰² )

8S = 3²⁰⁰⁴ - 3⁰ = 3²⁰⁰⁴ - 1

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

Answer 2

Nhân S với 3² ta có :

9S = 3² + 3⁴ + ... + 3²⁰⁰² + 3²⁰⁰⁴

9S - S = ( 3² + 3⁴ + ... + 3²⁰⁰⁴ ) - ( 3² + 3⁴ +... + 3²⁰⁰²)

8S = 3^2004-3 = 3 ( 3^{2003 - 1} )

=> 8S = \(\frac{3}{8}\). ( 3²⁰⁰³ - 1 )

k bít đúng k nữa , thầy cko lm mà thấy nó hơi khó hỉu pn , ráng hỉu nha =))

Answer 3

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

\(9S=9\left(3^0+3^2+3^4+3^6+................+3^{2002}\right)\)

\(9S=3^2+3^4+3^6+3^8+.............+3^{2004}\)

\(9S-S=\left(2^2+2^4+2^6+2^8+..........+2^{2004}\right)-\left(2^0+2^2+2^4+2^6+................+2^{2002}\right)\)\(8S=2^{2004}-2^0\)

\(8S=2^{2004}-1\)

\(S=\dfrac{2^{2004}-1}{8}\)