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The second day found regular math problems.
Pythagorean theorem This is just a general pattern.

3,4,5 3^2+4^2=5^2

5, 12, 13 5^2+ 12^2= 13^2

7,24,25 7^2+24^2=25^2

9,40,4 1 9^2+40^2=4 1^2

...........

2 1,a,b 2 1^2+b^2=C^2

(1) When a=2 1, the values of B and C.

(2) When a=2n+ 1, the answers of b and c are: 1. B = (square of 21-1) ÷ 2 = 220c = (square of 21+65438+.

2.b=[(2n+ 1) squared-1]÷2 c=[(2n+ 1) squared+1] ÷ 21* 3 *. (2) According to (1), calculate the result of 2000 * 2001* 2002 * 2003+1(expressed by a simplest formula) Answer:1) n (n+1. 5+3n+ 1)? 0? 5.

(n is a positive integer). The evidence is as follows:

n(n+ 1)(n+2)(n+3)+ 1

= n(n+3)(n+ 1)(n+2)+ 1

=(n? 0? 5+3n)(n? 0? 5+3n+2)+ 1

=(n? 0? 5+3n)? 0? 5+2(n? 0? 5+3n)+ 1

=(n? 0? 5+3n+ 1)? 0? 5.

(2)2000*200 1*2002*2003+ 1

=(2000? 0? 5+3*2000+ 1)? 0? five

= 400600 1? 0? There are several odd numbers starting from 1 on the blackboard. Erase an odd number and the sum of the remaining odd numbers is 2004. What is the erased odd number?

Answer: 2 1