Question

A= 3+3^2+3^3+......+3^99+3^100

Answer 1

Ta có: A=1+3+3^2+…+3^100 

=>A.3-A=3+3^2+3^3+…+3^101 -1-3-3^2 -…-3^100 

=>A.2=3^101 -1 

=>A.2+1=3^101=3^n 

=>3^101=3^n 

=>n=101 

Answer 2

A = 3 + 3² + 3³ + ... + 3⁹⁹ + 3¹⁰⁰

3A = 3² + 3³ + 3⁴ + ... + 3¹⁰⁰ + 3¹⁰¹

3A - A = ( 3² + 3³ + 3⁴ + ... + 3¹⁰⁰ + 3¹⁰¹ ) - ( 3 + 3² + 3³ + ... + 3⁹⁹ + 3¹⁰⁰ )

2A = 3¹⁰¹ - 3

A = \(\frac{3^{101}-3}{2}\)

Answer 3

\(A=3+3^2+3^3+....+3^{99}+3^{100}\)

\(A\times3=3+3^2+3^3+3^{101}\)

\(A\times3-A=3+3^2+3^3+...+3^{101}-1-3-3^2-...-3^{100}\)

\(A\times2=3^{101}-1\)

\(A\times2+1=3^{101}=3^n\)

\(N=101\)