Trả lời:

\(S=\) \(\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\)\(\left(-3\right)^{2015}\)

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

\(-3S-S=\)\([\left(-3\right)^1+\left(-3\right)^2+...+\)\(\left(-3\right)^{2016}\)\(]\)\(-\)\([\)\(\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\)\(\left(-3\right)^{2015}\)\(]\)

\(\left(-3-1\right)S=\)\(\left(-3\right)^1+\left(-3\right)^2+...+\)\(\left(-3\right)^{2016}\)\(-\)\(\left(-3\right)^0-\left(-3\right)^1-\left(-3\right)^2-...-\)\(\left(-3\right)^{2015}\)

\(-4S=\)\(\left[\left(-3\right)^1-\left(-3\right)^1\right]\)\(+\)\(\left[\left(-3\right)^2-\left(-3\right)^2\right]\)\(+\)\(...\)\(+\)\(\left[\left(-3\right)^{2015}-\left(-3\right)^{2015}\right]\)\(+\)\(\left[\left(-3\right)^{2016}-\left(-3\right)^0\right]\)

\(-4S=\)\(0+0+...+0+\left(-3\right)^{2016}-1\)

\(-4S=\)\(3^{2016}-1\)

\(S=\frac{-3^{2016}+1}{4}\)

Vậy \(S=\frac{-3^{2016}+1}{4}\)

P/s: Không chắc có đúng ko. 

Hok tốt!

Vuong Dong Yet

s = \(\left(-3\right)^{2016}-\left(-3\right)^0\)

a) \(S=1+\left(-2\right)+3+\left(-4\right)+...+\left(-2014\right)+2015\)

\(\Leftrightarrow S=\left(1-2\right)+\left(3-4\right)+....+\left(2013-2014\right)+2015\)

Vì từ 1 đến 2014 có 2014 số hạng => có 1007 cặp => Có 1007 cặp -1 và số 2015

\(\Rightarrow S=\left(-1\right)\cdot1007+2015\)

<=>S=-1007+2015

<=> S=1008

Ta có B= (-3)⁰+ (-3)¹+.....+(-3)²⁰¹⁵

=> -3B= -3.[(-3)⁰+(-3)¹+...+(-3)²⁰¹⁵]

=> -3B= (-3)¹+ (-3)²+....+(-3)²⁰¹⁶

=> -3B-B= (-3)¹ +(-3)²+....+ (-3)²⁰¹⁶ - [(-3)⁰+(-3)¹+....+ (-3) ²⁰¹⁵

=> -4B= (-3)²⁰¹⁶- (-3)¹

=>-4B= (-3)²⁰¹⁶+ 1

=> B= (-3)²⁰¹⁶+ 1 / -4

Mình nhầm, -4B= (-3)²⁰¹⁶- (-3)⁰

\(=\dfrac{\left(-5\right)\cdot0.81}{\left(\dfrac{5}{2}\right)^4\cdot\left(-\dfrac{10}{3}\right)^3\cdot\left(-1\right)}=\dfrac{-5\cdot0.81}{\dfrac{5^4\cdot10^3}{2^4\cdot3^3}}\)

\(=-4.05:\dfrac{5^7}{3^3\cdot2}=\dfrac{-81}{20}\cdot\dfrac{3^3\cdot2}{5^7}=\dfrac{-3^7\cdot2}{2^2\cdot5^8}=\dfrac{-3^7}{2\cdot5^8}\)