6)Prove that 1000^{2}-999^{2 }+998^{2}-997^{2}+...+2^{2}-1^{2} is equal to the sum of the first thousand positive integers

7) suppose that p>q>0. sketch a square of side p and another square of side q, and give a pictorial demonstration of the identity p^{2}-q^{2}= (p+q)(p-q)

8) verify that 13! = 78912^{2} -288^{2} without actually calculating the value of 13!

9) verify that 13! = 112296^{2}-79896^{2} without actually calculating the value of 13!

10) write 17! as the difference of two integer perfect squares

11) it is possible to arrive at the factorization of x^{2}-y^{2} by the technique of adding in an extra term and subtracting it out again:

x^{2}-y^{2}=x^{2}-xy+xy-y^{2}

=x(x-y)+y(x-y)

=(x+y)(x-y)

apply this technique to determine a factorization (with integer exponents) of x^{3 }-y^{3}

Guest Jan 1, 2019

#1**+2 **

6) This forms an arithemetic series as follows:

First term =3994, Number of term =[10,002 - 12 / 10 + 1] =1,000 / 4 =250, Common difference =-16

Sum = N/2[2F + D(N-1)

Sum =250/2[2*3,994 + (-16*(249)

Sum = 125 [7,988 - 3,984]

Sum =**500,500**

Sum of 1 to 1,000 =[1,000 x 1001] / 2 =**500,500**

Guest Jan 1, 2019

#2**+2 **

11. x^3 - y^3

x^3 - x^2 y + x^2y - xy^2 + xy^2 - y^3 =

x^2 ( x - y) + xy (x - y) + y^2 (x - y) =

(x - y) ( x^2 + xy + y^2)

CPhill Jan 2, 2019

#3**+2 **

8) verify that 13! = 78912^2 -288^2 without actually calculating the value of 13!

78912^2 - 288^2 =

(78912 + 288) (78912 - 288) =

(79200) (78624)

(2^5 * 3^2 * 5^2* 11 ) ( 2^5 * 3^3 * 7 * 13)

( 2 * 3 * 2^2 * 5 * ( 2 * 3) * 2 * 5 * 11) ( 2 * 2 * 2^3 * 3^2 * 3 * 7 * 13) =

( 2 * 3 * 4 * 5 * 6 ) ( 2 *5* 11) ( 7 * 2^3 * 3^2 ) ( 2 * 2 * 3 * 13 ) =

(2 * 3 * 4 * 5 * 6 * 7 * 8 * 9) ( (2 * 5) * 11 * ( 2 * 2 * 3) * 13 ) =

1 * (2 * 3 * 4 * 5 * 6 * 7 *8 * 9) ( 10 * 11 * 12 * 13 ) = 13!

CPhill Jan 2, 2019