Question

Tính C :

C=3+3^2+3^3+……..+3^100

Answer 1

\(C=3+3^2+3^3+...+3^{100}\)

\(\Rightarrow3C=3\left(3+3^2+3^3+...+3^{100}\right)\)

\(\Rightarrow3C=3^2+3^3+...+3^{101}\)

\(\Rightarrow3C-C=3^2+3^3+...+3^{101}-3-3^2-3^3-...-3^{100}\)

\(\Rightarrow2C=3^{101}-3\)

\(\Rightarrow C=\dfrac{3^{101}-3}{2}\)

Answer 2

C = 3 + 3² + 3³ + ... + 3¹⁰⁰
=> 3C = 3² + 3³ + ... + 3¹⁰⁰ + 3¹⁰¹
=> 3C - C = 3¹⁰¹ - 3
=> 2C = 3¹⁰¹ - 3
=> C = \(\dfrac{3^{101}-3}{2}\)
@Yuuki Tenpouin