Question

Cho A=3+3²+3³+...+3¹⁰⁰

Tính: 2⁵⁰(A.2+1)

Answer 1

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

\(3A-A=\left(3^2-3^2\right)+\left(3^3-3^3\right)+......+3^{101}-3\)

\(2A=3^{101}-3\)

A = \(\frac{3^{101}-3}{2}\)

\(2^{50}\left(A.2+1\right)=2^{50}.\left(\frac{3^{101}-3}{2}.2+1\right)=2^{50}.\left(3^{101}-2\right)\)

Answer 2

A = 3 + 3² + 3³ + ... + 3¹⁰⁰

3A = 3² + 3³ + ... + 3¹⁰¹

3A - A = 3¹⁰¹ - 3

2A = 3¹⁰¹ - 3

=> 2⁵⁰(3¹⁰¹ - 3 + 1 )

= 2⁵⁰.3¹⁰¹ - 2