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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1^2-2^2+3^2-4^2+...-2004^2+2005^24^{ }-4^2\)
\(=1^2+\left(3^2-2^2\right)+\left(5^2-4^2\right)+...+\left(2005^2-2014^2\right)\)
\(=1+5+9+...+4009\)
số số hạng có trong A là
\(\left(4009-1\right):4+1=1003\)
tổng cấc số hạng có trong A là
\(\left(4009+1\right).1003:2=2011015\)
vậy A = 2011015
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : A = (12 - 22) + (32 - 42) + .... + (20032 - 20042) + 20052
= (1 - 2)(1 + 2) + (3 - 4).(3 + 4) + .... + (2003 - 2004).(2003 + 2004) + 20052
= -1(1 + 2 + 3 + 4 + .... + 2003 + 2004) + 20052
= -1.2004.(2004 + 1) : 2 + 20052
= -1002.2005 + 2005.2005
= 2005.1003 = 2011015
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1^2-2^2+3^2-4^2+...-2004^2+2005^2\)
\(=1^2-2^2+3^2-4^2+...+2003^2-2004^2+2005^2\)
\(=-\left(2^2-1^2\right)-\left(4^2-3^2\right)-...-\left(2004^2-2003^2\right)+2005^2\)
\(=-\left[\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(2004^2-2003^2\right)\right]+2005^2\)
\(=-\left[\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(2004-2003\right)\left(2004+2003\right)\right]+2005^2\)
\(=-\left[1+2+3+4+...+2003+2004\right]+2005^2\)
\(=-\dfrac{2004.2005}{2}+2005^2\)
\(=2011015\)
:D
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(D=\dfrac{5^{4008}-5^{2004}-1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(A=1^2-2^2+3^2-4^2+...+2003^2-2004^2+2005^2\)
\(=-\left[2^2-1^2+4^2-3^2+...+2004^2-2003^2\right]+2005^2\)
\(=-\left[\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(2004-2003\right)\left(2004+2003\right)\right]+2005^2\)
\(=-\left(1+2+3+4+...+2003+2004\right)+2005^2\)
\(=-\dfrac{2005.2004}{2}+2005^2\)
\(=6029035\)
P/s : Làm linh tinh , ko chắc :
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+...+\left(2003-2004\right)\left(2003+2004\right)+2005^2\)
\(=2005^2-\left(1+2+3+...+2004\right)\)
=2005^2-2009010
=2011015
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt dãy trên là A
Ta có:
A=(12-22)+(32-42)+...+(20032-20042)+20052
A=(1-2)(1+2)+(3-4)(3+4)+...+(2003-2004)(2003+2004)+20052
A=(-1.3)+(-1.7)+(-1.11)+...+(-1.4007)+4020025
A=-3+(-7)+(-11)+...+(-4007)+4020025
A=-(3+7+11+...+4007)+4020025
A=-{(4007+3)[(4007-3):4+1]}+4020025
A=-(4010.1002)+4020025
A=-4018020+4020025
A=2005
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