Question

tính S=3^0+3^2+3^4+3^6+...+3^2002

 

 

Answer 1

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

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

\(3S-S=\left(3^2+3^4+3^6+...+3^{2004}\right)-\left(3^0+3^2+3^4+....+3^{2002}\right)\)

\(3S-S=3^{2004}-3^0\)

\(S=\frac{3^{2004}-3^0}{2}\)

Answer 2

S = 3⁰ + 3² + 3⁴ + .... + 3²⁰⁰² 

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

= 3² + 3⁴ + 3⁶ + .... + 3²⁰⁰⁴

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

8S = 3²⁰⁰⁴ - 1

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