Có B = 1-3+\(3^2-3^3+...+3^{2014}-3^{2015}\)
3B = 3.(1-3+\(3^2-3^3+...+3^{2014}-3^{2015}\))
3B = 3\(-3^2+3^3-3^4+...+3^{2015}-3^{2016}\)
3B+B = (3\(-3^2+3^3-3^4+...+3^{2015}-3^{2016}\))+(1-3+\(3^2-3^3+...+3^{2014}-3^{2015}\))
4B = 1\(-3^{2016}\) => B = \(\left(1-3^{2016}\right)\div4\) B = \(\dfrac{1}{4}-\dfrac{3^{2016}}{4}\)<\(\dfrac{1}{4}\) (đpcm)