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Tính tổng
S=\(\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+........+\left(-3\right)^{2015}\)
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
Ta có :
\(S=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{2015}\)
\(3S=\left(-3\right)^1+\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^{2015}\)
\(3S-S=\left[\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{2016}\right]+\left[\left(-3\right)^0+\left(-3\right)^1+...+\left(-3\right)^{2015}\right]\)
\(2S=\left(-3\right)^{2016}-\left(-3\right)^0\)
\(2S=3^{2016}-1\)
\(S=\frac{3^{2016}-1}{2}\)
Vậy \(S=\frac{3^{2016}-1}{2}\)
Chúc bạn học tốt ~
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
\(\left(-3\right)\cdot B=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{2016}\)
=>-4B=(-3)^2016-1
=>\(B=\dfrac{-3^{2016}+1}{4}\)
\(=\dfrac{\left(-5\right)\cdot0.81}{\left(\dfrac{5}{2}\right)^4\cdot\left(-\dfrac{10}{3}\right)^3\cdot\left(-1\right)}=\dfrac{-5\cdot0.81}{\dfrac{5^4\cdot10^3}{2^4\cdot3^3}}\)
\(=-4.05:\dfrac{5^7}{3^3\cdot2}=\dfrac{-81}{20}\cdot\dfrac{3^3\cdot2}{5^7}=\dfrac{-3^7\cdot2}{2^2\cdot5^8}=\dfrac{-3^7}{2\cdot5^8}\)
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
