Question

tính tổng S = 1 + 3¹ + 3² + 3³ + ..... + 3¹⁰¹

Answer 1

`#3107.101107`

\(S=1+3^1+3^2+3^3+...+3^{101}\)

\(3S=3+3^2+3^3+...+3^{102}\)

\(3S-S=\left(3+3^2+3^3+...+3^{102}\right)-\left(1+3+3^2+...+3^{101}\right)\)

\(2S=3+3^2+3^3+3^{102}-1-3-3^2-...-3^{101}\)

\(2S=3^{102}-1\)

\(S=\dfrac{3^{102}-1}{2}\)

Vậy, \(S=\dfrac{3^{102}-1}{2}.\)

Answer 2

3s=3+3^2+3^3+....+3^102

3s-s=2s

2s=3^102-1

s=3^102-1 trên2