Question

[Nam Hạo Dương]

Tính tổng sau:

C=3^2+3^4+3^6+....+3^202+3^204

D=1/3^2-1/3^4+1/3^6-1/3^8+......+1/3^202-1/3^204

Answer 1

[Xyz OLM]

a) Ta có : C = 3² + 3⁴ + 3⁶ + ... + 3²⁰² + 3²⁰⁴

=> 3²C = 9C = 3⁴ + 3⁶ + 3⁸ + .... + 3²⁰⁴ + 3²⁰⁶

Lấy 9C trừ C theo vế ta có 

9C - C = (3⁴ + 3⁶ + 3⁸ + .... + 3²⁰⁴ + 3²⁰⁶) - ( 3² + 3⁴ + 3⁶ + ... + 3²⁰² + 3²⁰⁴)

=> 8C = 3²⁰⁶ - 3²

=> C = \(\frac{3^{206}-3^2}{8}\)

d) Ta có D = \(\frac{1}{3^2}-\frac{1}{3^4}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\)

=> 3²D = 9D = \(1-\frac{1}{3^2}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\)

Lấy 9D cộng D theo vế ta có :

9D + D = \(\left(1-\frac{1}{3^2}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\right)+\left(\frac{1}{3^2}-\frac{1}{3^4}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)

=> 10D = \(1-\frac{1}{3^{204}}\)

=> D = \(\frac{1}{10}-\frac{1}{3^{204}.10}\)