Question

Bài 5:

Answer 1

\(a,B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\\ B=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{99}\left(1+3\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{99}\right)=4\left(3+3^3+...+3^{99}\right)⋮2\left(4⋮2\right)\\ b,3B=3^2+3^3+...+3^{101}\\ \Rightarrow3B-B=3^2+3^3+...+3^{101}-3-3^2-3^3-...-3^{100}\\ \Rightarrow2B=3^{101}-3\\ \Rightarrow B=\dfrac{3^{101}-3}{2}\)