Question

B=1+3+3^2+........+3^100

Answer 1

B = 1+3+3²+......+3¹⁰⁰

=> 3B = 3+3²+3³+...+3¹⁰¹

=> 3B-B = 3+3²+3³+...+3¹⁰¹ - ( 1+3+3²+......+3¹⁰⁰ ) 

=> 2B = 3¹⁰¹-1

=> B = \(\frac{3^{101}-1}{2}\)

Answer 2

\(B=1+3+3^2+...+3^{100}\)

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

\(3B-B=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)

\(2B=3^{101}-1\)

\(B=\frac{3^{101}-1}{2}\)

Ủng hộ mk nha ^-^

Answer 3

\(B=1+3+3^2+...+3^{100}\)

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

\(3B-B=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)

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

\(2B=3^{101}-1\)

\(B=\frac{3^{101}-1}{2}\)