Question

A=1+3+3²+3³+....+3³³

\(B=\frac{3^{21}}{2}\)

Tính B-A

Answer 1

A=1+3+3²+3³+....+3³³

3A=3+3²+3³+3⁴+...+3³⁴

2A=3A-A=( 3+3²+3³+3⁴+...+3³⁴ ) - ( 1+3+3²+3³+....+3³³ )

2A=3³⁴ - 1

A = \(\frac{3^{34}-1}{2}\)

B-A=\(\frac{3^{21}-3^{34}+1}{2}\)

Answer 2

giúp mình với

Answer 3

=\(3^0+3^1+...+3^{33}\)

3A=\(3^1+3^2+....+3^{34}\)

3A-A=(\(3^1+3^2+....+3^{34}\))-(\(3^0+3^1+...+3^{33}\))

2A=3^34-1

A=(3^34-1):2

\(\frac{3^{21}}{2}-\frac{3^{24}}{2}\)