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S=1-3+32-33+...+32014-32015
=>3S=3-32+...+32015-32016
=>3S+S=4S=(3-32+...+32015-32016)+(1-3+...+32014-32015)
=>4S=-32016+1
=>S=\(-\frac{3^{2016}-1}{4}\)
\(S=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+\left(-3\right)^3+........+\left(-3\right)^{2015}\)
\(\Rightarrow-3S=\left(-3\right)^1+\left(-3\right)^2+\left(-3\right)^3+\left(-3\right)^4+......+\left(-3\right)^{2016}\)
\(\Rightarrow-4S=\left[\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{2016}\right]-\left[\left(-3\right)^0+\left(-3\right)^1+...+\left(-3\right)^{2015}\right]\)
\(\Rightarrow-4S=\left(-3\right)^{2016}-\left(-3\right)^0\Rightarrow-4S=3^{2016}-1\Rightarrow S=\frac{3^{2016}-1}{-4}\)
Ta có :
\(S=\left(-3\right)^0+\left(-3\right)+\left(-3\right)^2+..................+\left(-3\right)^{2015}\)
\(\Rightarrow\left(-3\right).S=\left(-3\right)+\left(-3\right)^2+\left(-3\right)^3+..............+\left(-3\right)^{2015}+\left(-3\right)^{2016}\)
\(\Rightarrow\left(-3\right).S-S=\left[\left(-3\right)+\left(-3\right)^2+..............+\left(-3^{2015}\right)+\left(-3\right)^{2016}\right]-\left[\left(-3\right)^0+\left(-3\right)+...........+\left(-3\right)^{2015}\right]\)\(\Rightarrow\left(-4\right)S=\left(-3\right)^{2016}-\left(-3\right)^0\)
\(\Rightarrow\left(-4\right).S=\left(-3\right)^{2016}-1\)
\(\Rightarrow S=\dfrac{\left(-3\right)^{2016}-1}{-4}\)
\(\Rightarrow S=\dfrac{3^{2016}-1}{-4}\)
Đặt \(A=1+3+3^2+3^3+....+3^{2015}-3^{2016}\)
\(B=1+3+3^2+3^3+....+3^{2015}\)
Ta có:
\(B=1+3+3^2+...+3^{2015}\)
\(\Rightarrow3B=3+3^2+3^3+...+3^{2016}\)
\(\Rightarrow3B-B=\left(3+3^2+3^3+...+3^{2016}\right)-\left(1+3+3^2+...+3^{2015}\right)\)
\(\Rightarrow2B=3^{2016}-1\)
\(\Rightarrow B=\frac{3^{2016}-1}{2}\)
\(\Rightarrow A=\frac{3^{2016}-1}{2}-3^{2016}\)
S = (-3)0 + (-3)1 + (-3)2 + ... + (-3)2015
=> -3S = (-3)1 + (-3)2 + ... + (-3)2016
=> -4S = (-3)2016 - 1
=> S = \(\dfrac{3^{2016}+1}{4}\)
32010- ( 32009 + 32008 + ... + 3 + 1 )
Đặt A = 1 + 3 + ... + 32009
=> 3A = 3 + 32 + ... + 32010
=> 3A - A = 32010 - 1
Nên 32010 - ( 32010 - 1 ) = 1
Mình nhầm xíu :
Tính giá trị của biểu thức :
P = x2015 + y2015 + z2015
a) Đặt \(A=2^{2016}-2^{2015}+2^{2014}-2^{2013}+...+2^2-2^1\)
\(\Rightarrow2A=2^{2017}-2^{2016}+2^{2015}-2^{2014}+...+2^3-2^2\)
\(\Rightarrow2A+A=\left(2^{2017}-2^{2015}+2^{2014}-2^{2013}+...+2^3-2^2\right)+\left(2^{2016}-2^{2015}+2^{2014}-2^{2013}+...+2^2+2^1\right)\)
\(\Rightarrow3A=2^{2017}+1\)
\(\Rightarrow A=\frac{2^{2017}+1}{3}\)
b) Đặt \(B=3^{1000}-3^{999}+3^{998}-3^{997}+...+3^2-3^1+3^0\)
\(\Rightarrow3B=3^{1001}-3^{1000}+3^{999}-3^{997}+...+3^3-3^2+3^1\)
\(\Rightarrow3B+B=\left(3^{1001}-3^{1000}+3^{999}-3^{998}+...+3^3-3^2+3^1\right)+\left(3^{1000}-3^{999}+3^{998}-3^{997}+...+3^2-3^1+3^0\right)\)
\(\Rightarrow4B=3^{1001}+3^0\)
\(\Rightarrow B=\frac{3^{1001}+1}{4}\)
a) Đặt A = 22016 - 22015 + 22014 - 22013 + ... + 22 - 21
2A = 22017 - 22016 + 22015 - 22014 + ... + 23 - 22
2A + A = (22017 - 22016 + 22015 - 22014 + ... + 23 - 22) + (22016 - 22015 + 22014 - 22013 + ... + 22 - 21)
3A = 22017 - 21
3A = 22017 - 2
\(A=\frac{2^{2017}-2}{3}\)
b) lm tương tự câu a
S = (-3)0 + (-3)1 + (-3)2 + (-3)3 +......+ (-3)2015
=>-3S= (-3)1 + (-3)2 + (-3)3 +......+ (-3)2015+(-3)2016
=>-3S-S=[ (-3)1 + (-3)2 + (-3)3 +......+ (-3)2015+(-3)2016]-[ (-3)0 + (-3)1 + (-3)2 + (-3)3 +......+ (-3)2015]
=>-4S=(-3)1 + (-3)2 + (-3)3 +......+ (-3)2015+(-3)2016 -(-3)0 - (-3)1 - (-3)2 - (-3)3 -......- (-3)2015
=>-4S=(-3)2016-(-3)0
=>-4S=(-3)2016-1
=>S=\(\frac{\left(-3\right)^{2016}-1}{-4}=\frac{3^{2016}-1}{-4}\)