(n+ 1)！ -No! = n*n！

n(n+ 1)！ - n*n！ = n^2*n！

= & gtn^2*n！ + n*n！ + n！ =

n(n+ 1)！ - n*n！ + (n+ 1)！ -No! + n！ = (n+ 1)(n+ 1)！ - n*n！

Original title s+1=101*10/! - 100 * 100! + 100* 100! - 99*99! +...+ 2*2! - 1* 1! + 1 = 10 1* 10 1!

So the result is101*10/! / 10 1! = 10 1

Square number problem:

Let m = a 2+b 2 and n = X 2+Y 2, then Mn = (A2+B2) (X2+Y2) = (AX) 2+(BY) 2+(AY) 2+(BX) 2 = (AX+BY) 2-2. 2 It can be known that mn is still of "1" type. (2) If M = A 2+KB 2 N = X 2+KY 2, then Mn = (AX) 2+(KBY) 2+K (BX) 2+K (AY) 2 = (.

Supplement:

The second problem, xy+2zw = 7xz-yw = 3, can be handled by the above type 2 number, and 2 comes from 2zw, that is, Mn = (xy+2zw) 2+2 * (zx-yw) 2 = 7 2+2 * 3 2 = 67mn is a prime number, so that m = 65437. X = 1, w = 0 or x=-1, w = 0 = & gtY= 7, z = 3, y = 7, z = -3 or y=-1 z = 3, y=-1. A combination (x, y, z, w) that satisfies the condition (1, 7, 3, 0).

10. Triangle area

Make a parallelogram through the midpoint of one side, and find the dividing line of the two middle lines as follows.

2 3

6 4

The area of the ladder is 3/4 of that of a triangle (because triangles are similar).

Let's find the trapezoidal area.

12 Sina+3 Sina+6 sinb+6 sinb(a+b = 180)Sina = sinb

15 Sina+12 Sina = 27 Sina

Maximum when α= 90 degrees.

So the maximum area of a triangle is 27 * 4/3 = 36.