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Đặt A=1+2-3-4+5-6-7-8+9+10-...+2006-2007-2008+2009
Ta có:A=1+2-3-4+5+6-7-8+9+10-...+2006-2007-2008+2009
A=1+(2-3-4+5)+(6-7-8+9)+...+(2006-2007-2008+2009)
A=1+0+0+0+....+0
A=1
= 1+(2-3-4+5)+(6-7-8+9)+ ... +(2006-2007-2008+2009)
=1+0+0+ ... +0
=1
E = 1 - 3 + 5 - 7 + ... + 2013 - 2015 (1008 số hạng)
= (1 - 3) + (5 - 7) + ... + (2013 - 2015) (504 cặp)
= -2 + (-2) + ... + (-2) (504 số hạng - 2)
= 504.(-2) = -1008
F = 1 + 2 - 3 - 4 + .... + 2006 - 2007 - 2008 + 2009 (2009 SỐ HẠNG)
= 1 + (2 - 3) + (-4 + 5) + .... + (2006 - 2007) + (-2008 + 2009) (1004 cặp số)
= 1 + [(-1) + 1 + .... + (-1) + 1] (1005 số hạng)
= 1
E = 1 - 3 + 5 - 7 + .. + 2013 - 2015
E= ( 1 - 3 ) + ( 5 - 7 ) + ... + ( 2013 - 2015 )
E = - 2 + ( -2 ) + ... + (-2 )
E = -2 x 504
E= -1008
a) \(1+2+3+4+...+n\)
\(=\left(n+1\right)\left[\left(n-1\right):1+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right):2\)
\(=n\left(n+1\right):2\)
\(=\dfrac{n\left(n+1\right)}{2}\)
b) \(2+4+6+..+2n\)
\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)
\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right)\)
\(=n\left(n+1\right)\)
c) \(1+3+5+...+\left(2n+1\right)\)
\(=\left[\left(2n+1\right)+1\right]\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}:2\)
\(=\left(2n+1+1\right)\left[\left(2n-1-1\right):2+1\right]:2\)
\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)
\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right)\)
\(=n\left(n+1\right)\)
d) \(1+4+7+10+...+2005\)
\(=\left(2005+1\right)\left[\left(2005-1\right):3+1\right]:2\)
\(=2006\cdot\left(2004:3+1\right):2\)
\(=2006\cdot\left(668+1\right):2\)
\(=1003\cdot669\)
\(=671007\)
e) \(2+5+8+...+2006\)
\(=\left(2006+2\right)\left[\left(2006-2\right):3+1\right]:2\)
\(=2008\cdot\left(2004:3+1\right):2\)
\(=1004\cdot\left(668+1\right)\)
\(=1004\cdot669\)
\(=671676\)
g) \(1+5+9+...+2001\)
\(=\left(2001+1\right)\left[\left(2001-1\right):4+1\right]:2\)
\(=2002\cdot\left(2000:4+1\right):2\)
\(=1001\cdot\left(500+1\right)\)
\(=1001\cdot501\)
\(=501501\)
=(1+2-3-4)+(5+6-7-8)+...+(2001+2002-2003-2004)+2005+2006
=-4+(-4)+(-4)+...+(-4)+2005+2006
=501(-4) +2005+2006
=-2004 +2005+2006
=2007
1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - ... + 2006
= 1 + ( 2 - 3 - 4 + 5 ) + ( 6 - 7 - 8 + 9 ) + ( 10 - 11 - 12 + 13 ) + ... + ( 2002 - 2003 - 2004 + 2005 ) + 2006
= 1 + 0 + 0 + 0 + ... + 0 + 2006
= 1 + 2006
= 2007
chuk hok tốt