Theo nhị thức Newton ta có:

( 3 + x ) 2015 = ​​​ C 2015 0 .    3 2015 + ​   C 2015 1 .3 2014 . x + ​ C 2015 2 .3 2013 . x 2 + ....   + C 2015 2014 .3. x 2014 + ​ C 2015 2015 . x 2015

Thay x = -1 ta được:

( 3 − 1 ) 2015 = ​​​ C 2015 0 .    3 2015 − ​   C 2015 1 .3 2014 + ​ C 2015 2 .3 2013 − ....   + C 2015 2014 .3 − ​ C 2015 2015

Suy ra,  S   =     2 2015

 Ta chọn đáp án A

\(B=1+3^1+3^2+...+3^{2016}\)

\(3B=3+3^2+3^3+3^4+...+3^{2017}\)

\(3B-B=3^{2017}-1\)

\(B=\dfrac{3^{2017}-1}{2}\)

\(B=1+3^1+3^2+...+3^{2016}\)

\(3\cdot B=3+3^2+3^3+...+3^{2016}+3^{2017}\)

\(3B-B=3+3^2+3^3+...+3^{2016}+3^{2017}-\left(1+3^1+3^2+...+3^{2016}\right)\)

\(2B=3^{2017}-1\)

\(\Rightarrow B=\dfrac{3^{2017}-1}{2}\)

\(B=1+3+3^2+...+3^{2016}\)

\(3\cdot B=3+3^2+3^3+...+3^{2017}\)

\(3B-B=3+3^2+3^3+...+3^{2017}-\left(1+3+3^2+...+3^{2016}\right)\)

\(2B=3^{2017}-1\)

\(\Rightarrow B=\dfrac{3^{2017}-1}{2}\)

42016 x 3 - 32017=126048-32017

=94030

k nha 

Huy Rio ơi đơn vị bạn tính sai rùi đáp án là 94031 mới đúng

\(A=3+3^2+3^3+...+3^{2015}\)

\(\Rightarrow3A=3^2+3^3+...+3^{2015}+3^{2016}\)

\(\Rightarrow3A-A=\left(3^2+3^3+...+3^{2016}\right)-\left(3+3^2+3^3+...+3^{2015}\right)\)

\(\Rightarrow2A=\left(3^2-3^2\right)+\left(3^3-3^3\right)+...+\left(3^{2016}-3\right)\)

\(\Rightarrow2A=3^{2016}-3\)

\(\Rightarrow A=\dfrac{3^{2016}-3}{2}\)

Ta có: \(2A+3=3^n\)

\(\Rightarrow2\cdot\dfrac{3^{2016}-3}{2}+3=3^n\)

\(\Rightarrow3^{2016}-3+3=3^n\)

\(\Rightarrow3^{2016}=3^n\)

\(\Rightarrow n=2016\)