Question

 

Tính tổng :
Q= 1+3 mũ 3 + 3 mũ 6+ 3 mũ 9+.....+ 3 mũ 99

Answer 1

\(Q=1+3^3+3^6+3^9+...+3^{99}\)

\(3^3Q=3^3+3^6+3^9+3^{12}+...+3^{102}\)

\(27Q-Q=\left(3^3+3^6+3^9+3^{12}+...+3^{102}\right)-\left(1+3^3+3^6+3^9+...+3^{99}\right)\)

\(26Q=3^{102}-1\)

\(Q=\dfrac{3^{102}-1}{26}\)

Answer 2

     Q =  1 + 3³ + 3⁶ + 3⁹ + ... + 3⁹⁹

3³Q =   3³ + 3⁶ + 3⁹ + 3¹²+ ...+ 3¹⁰²

27Q - Q = 3³ + 3⁶ + 3⁹ + 3¹² +... + 3¹⁰² - (1 + 3³ + 3⁶ + 3⁹ + ... + 3⁹⁹)

26Q = 3³ + 3⁶ + 3⁹ + 3¹² + ... + 3¹⁰² - 1 - 3³ - 3⁶ - 3⁹ - ...- 3⁹⁹

26Q = (3¹⁰² - 1) + (3³ - 3³) + (3⁶ - 3⁶) + ... + (3⁹⁹ - 3⁹⁹)

26Q = 3¹⁰² - 1 + 0  + 0 + 0 + ... + 0

26Q = 3¹⁰² - 1

Q = \(\dfrac{3^{102}-1}{26}\)