2005x2004-2004x2003+2003x2002-2002x2002+...+3x2-2x1=

2005x2004-2004x2003+2003x2002-2002x2002+...+3x2-2x1=

Every two multipliers are combined to get: (2005-2003) x2004 + (2003-2001) x2002 +. (3-1) x2 check again: 2x2004 + 2x2002 + 2x2000 +. 2x2 continue: 2x (2004 + 2002 + 20000 +. + 2) continue: 2 to 2004, add the first and last in turn: 2004 + 2 = 20042002 + 4 = 2006,. 1004 + 1002 = 2

How to do 2005x2004-2004x2003-2003x2002-2002x2001 +... + 3x2-2x1
How to solve it in a simple way,

2005x2004-2004x2003-2003x2002-2002x2001 +... + 3x2-2x1 = 2004 × (2005-2003) + n × (n + 1-N + 1) + (n-2) (n-2 + 1-N + 2 + 1)... 2 × (3-1) = 2004 × 2 + 2n + 2 (n-2)... 2 × 2n = 2004 / 2 = 1002 is calculated as 2S (1002) = 2 × (2 + 2004) × 1002 △ 2 =

The relationship between y = 3x2-1 and y = 3 {X-1} 2 and y = 3x2
Right now

Image position relationship between y = 3x ^ 2-1 and y = 3 (x-1) ^ 2 and y = 3x ^ 2
The image of y = 3x ^ 2-1 is obtained by moving the image of y = 3x ^ 2 down one unit;
The image of y = 3 (x-1) ^ 2 is obtained by moving the image of y = 3x ^ 2 one unit to the right

3x2-2x = 2, find 2 | 3x2-x + 1

3x & # 178; - 2x = 2, find 3x & # 178; - x + 1?
3x²-2x=2
Divide both sides by 2:
(3/2)x²-x=1;
2 / 3x & # 178; - x + 1
=[(3/2)x²-x]+1
=1+1
=2

Calculation: (3x2-2y) (- 3x2-2y)=

:(3x2-2y)(-3x2-2y)
=（-2y+3x^2)(-2y-3x^2)
=4y^2-9x^4

2008+2007+2006… +3+2+1

Starting from 1, if the mantissa is double, take its first digit multiplied by half of the total number. If the mantissa is singular, multiply the last digit by half of the total number plus 1

How to do (1 + 3 + 5 +... 2007) - (2 + 4 + 6 +... + 2006)?

[(1+2007)*1004]/2-[(2+2006.*1003)]/2
This is the problem of summation of arithmetic sequence. Just add the first term and the last term, multiply by the number of terms, divide by two, and then subtract the two sides

1+2+2^2+2^3+2^4+.+2^2006+2^2007=

Because 1 + 2 = 2 ^ 0 + 2 ^ 1, 1 + 2 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 +. + 2 ^ 2006 + 2 ^ 2007 = 2 ^ 0 + 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 +. + 2 ^ 2006 + 2 ^ 2007,
At this time, we need to use the formula of equal ratio sequence = 2 ^ 2008-1

(1-1/2^2)(1-1/3^2)(1-1/4^2).(1-1/2006^2)(1-1/2007^2)=

Using the formula of square difference
The original formula = (1-1 / 2) (1 + 1 / 2) (1-1 / 3) (1 + 1 / 3) (1-1/2006)(1+1/2006)(1-1/2007)(1+1/2007)
=(1/2)(3/2)(2/3)(4/3)…… (2005/2006)/(2007/2006)(2006/2007)(2008/2007)
Middle cut
=(1/2)(2008/2007)
=1004/2007

1/1*2*3+1/2*3*4.1/2005*2006*2007

1/1*2*3+1/2*3*4.1/2005*2006*2007=1/2*(1/1*2-1/2*3)+1/2*(1/2*3-1/3*4)+1/2*(1/3*4-1/4*5)+.+1/2*(1/2005*2006-1/2006*2007)=1/2*(1/1*2-1/2*3+1/2*3-1/3*4+1/3*4-1/4*5+...+/2005*2006-1/2006*2007)=1/2*(1/2-1/2...