nice math riddle

2005! - 1

SUM(i=1 to n, i*i!) = (n+1)! - 1

Inductive proof:

True for n=1:

1*1! = 1 = (1+1)! - 1

Assume true for n, show true for n+1:
Need to show:
SUM(i=1 to n+1, i*i!) = (n+2)! - 1

SUM(i=1 to n+1, i*i!) =
SUM(i=1 to n), i*i!) + (n+1)(n+1)! =
(n+1)! - 1 + (n+1)(n+1)! =
(n+1)!(n+1+1) - 1 =
(n+1)!(n+2) - 1 =
(n+2)! - 1

QED

good job, you are correct! another riddle in math. there is a circle o that its radius is r. it's divided to 3 equal parts/arch (a,b,c). around these points (a,b,c) there are arch with radius r inside the circle. express the rose space that was created. the picture :

The area of the rose is equivalent to the whole circle minus the regular inscribed hexagon.

i.e. piR^2 - 3R^2 sin60
= (pi - 1.5 sqrt3)R^2

can you show me where is the inscribed hexagon?

The inscribed hexagon is the one generated by a compass stepping round the circle giving six equal arcs. i.e. it is composed of six equilateral triangles side R (triangle area = 1/2 R x R x sin60). Each triangle leaves an area between itself and the circumference equal to half a "rose petal".

good job! new riddle :
ab+ac+bc=1
a,b,c<=1
proof : a^2+b^2+c^2<=2

Are you missing a constraint? a, b, c >= 0 or |a|, |b|, |c| <= 1

Otherwise, a = -2, b = -2, c = 0.75 yields a^2 + b^2 + c^2 = 8.5625

you're right, my bad! |a|,|b|,|c|<=1

http://www.physicsforums.com/archive/topic/t-56251_a_problem.html

The problem statement is slightly different (a, b, c>= 0). However, it doesn't make a difference.

All three negative is equivalent to all three positive.
One or two negative makes it impossible to have ab+ac+bc=1 since two terms would be negative.