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![](https://rs.olm.vn/images/avt/0.png?1311)
![](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
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(K=1^2-2^2+3^2-4^2+......+2005^2\)
\(\Rightarrow K=1^2+\left(3^2-2^2\right)+\left(5^2-4^2\right)+.....\) \(+\left(2005^2-2004^2\right)\)
\(=1+\left(3-2\right)\left(3+2\right)+\left(5-4\right)\left(5+4\right)\)\(+......+\left(2005-2004\right)\left(2005+2004\right)\)
\(\Rightarrow K=1+5+9+13+.....+4009\)
Số số hạng trong tổng K là \(\frac{4009-1}{4}+1=1003\)
\(\Rightarrow K=\frac{\left(4009+1\right).1003}{2}=2005.1003\) = 2011015
![](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 12 - 22 = - 3
32 - 42 = - 7
.................
20052 - 20062 = -4011
-{(4011+3)[(4011-3):4+1]:2} = -2013021
![](https://rs.olm.vn/images/avt/0.png?1311)
7) \(A=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)
\(A=\left(-1\right)\left(1^{ }+2\right)+\left(-1\right)\left(3+4\right)+...+\left(-1\right)\left(2003+2004\right)+2005^2\)
\(A=-\left(1+2+3+...+2004\right)+2005^2\)
\(A=-\dfrac{2004.\left(2004+1\right)}{2}+2005^2\)
\(A=-1002.2005+2005^2\)
\(A=2005\left(2005-1002\right)=2005.1003=2011015\)
8) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\dfrac{\left(2^2-1\right)}{2-1}\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{64}-1\right)-2^{64}\)
\(B=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](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)
\(A=\frac{2004^3+1}{2004^2-2003}\)
\(A=\frac{2004+1}{1-2003}\)\(=\frac{2005}{-2002}\)
\(B=\frac{2005^3-1}{2005^2+2006}\)\(=\frac{2005-1}{1+2006}=\frac{2004}{2007}\)
\(\Rightarrow A>B\)
\(A=\frac{2004^3+1}{2004^2-2003}\)
\(A=\frac{\left(2004+1\right)\left(2004^2-2004+1\right)}{2004^2-2003}\)
\(A=\frac{2005.\left(2004^2-2003\right)}{2004^2-2003}=2005\)
\(B=\frac{2005^3-1}{2005^2+2006}\)
\(B=\frac{\left(2005-1\right)\left(2005^2+2005+1\right)}{2005^2+2006}=\frac{2004.\left(2005^2+2006\right)}{2005^2+2006}=2004\)
Tham khảo nhé~
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