Question

Cho B=3+3²+3³+...+3¹⁰⁰.Tìm số tự nhiên n, biết 2B+3=3ⁿ

Answer 1

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

\(=>3B=3^2+3^3+...+3^{100}+3^{101}\)

\(3B-B=\left(3^2+3^3+...+3^{100}+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)

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

Ta có: \(3^{101}-3+3=3^n\)

\(=>3^{101}=3^n\)

\(n=101\)

Answer 2

ta có:

3b= 3^2+3^3+3^4+.......+3^101

3b-b= 3^101-3

vậy 3^n=101

Answer 3

101 la dung

nho k nhe

chuc ban hoc gioi