a, \(A=-1^2+2^2-3^2+4^2-...-2017^2+2018^2\)
\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(2018^2-2017^2\right)\)
\(=\left(1+2\right)\left(2-1\right)+\left(3+4\right)\left(4-3\right)+...+\left(2017+2018\right)\left(2018-2017\right)\)
\(=1+2+3+4+...+2017+2018\)
\(=\dfrac{\left(2018+1\right).2018}{2}=2037171\)
Vậy A=2037171
b, \(B=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)
\(=-\left[\left(2^2-1^2\right)+\left(4^2-3^2\right)+...\left(2004^2-2003^2\right)\right]+2005^2\)
\(=-\left[\left(1+2\right)\left(2-1\right)+\left(3+4\right)\left(4-3\right)+...+\left(2003+2004\right)\left(2004-2003\right)\right]+2005^2\)
\(=-\left(1+2+3+4+...+2004\right)+2005^2\)
\(=-\dfrac{2005.2004}{2}+2005^2=-2009010+4020025\)
\(=2011015\). Vậy B=2011015
c, \(C=\left(2+1\right)\left(2^2+1\right)...\left(2^{128}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{128}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)...\left(2^{128}+1\right)\)\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{128}+1\right)\)
...
\(=\left(2^{128}-1\right)\left(2^{128}+1\right)=2^{256}-1\)
Vậy \(C=2^{256}-1\)
d, \(D=\left(5+1\right)\left(5^2+1\right)...\left(5^{2004}+1\right)-5^{2008}\)
\(\Rightarrow4D=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)...\left(5^{2004}+1\right)-5^{2008}\)
\(=\left(5^2-1\right)\left(5^2+1\right)...\left(5^{2004}+1\right)-5^{2008}\)
\(=\left(5^4-1\right)\left(5^4+1\right)...\left(5^{2004}+1\right)-5^{2008}\)
...
\(=\left(5^{2004}-1\right)\left(5^{2004}+1\right)-5^{2008}\)
\(=5^{4008}-1-5^{2008}\Rightarrow D=\dfrac{5^{4008}-5^{2008}-1}{4}\)
Vậy \(D=\dfrac{5^{4008}-5^{2004}-1}{4}\)