MATH
posted by Gerald .
1. Let triangle ABC be a triangle such that angle ACB is 135 degrees.
Prove that AB^2 = AC^2 + BC^2  (Root 2) x AC x BC
2. Simplify:
2/1!  3/2! + 4/3!  5/4! + ..... + 2010/2009!  2011/2010!
Where k! is the multiplication of all numbers up to k (e.g. 5! = 5x4x3x2x1=120)
3. For the algebra nerds:
Factorise:
a^4 + b^4 + c^2  2 (a^2 x b^2 + a^2 x c + b^2 x c)

1. Let triangle ABC be a triangle such that angle ACB is 135 degrees.
Prove that AB^2 = AC^2 + BC^2  (Root 2) x AC x BC
There is probably an error in the sign, because the result is obtained directly by application of the cosine law, and substituting cos(135°)=√2
AB^2 = AC^2 + BC^2  2 AC x BC cos(∠ACB)
=AC^2 + BC^2 + 2√2 AC x BC
So there must have been a misprint in the original question.
2. Simplify:
2/1!  3/2! + 4/3!  5/4! + ..... + 2010/2009!  2011/2010!
Where k! is the multiplication of all numbers up to k (e.g. 5! = 5x4x3x2x1=120)
With a little patience, it can be shown that the sum of the sequence having the general term k:
(1)^k * k/(k1)!
for k=2 to n is
((n1)!+(1)^k)/(k1)!
Thus for n=2011,
the sum is (2010!+1)/2010!
3. For the algebra nerds:
Factorise:
a^4 + b^4 + c^2  2 (a^2 x b^2 + a^2 x c + b^2 x c)
The expression factorizes to
Rearrange the expression to
a^4+b^4+2a²b²+c²2a²c2b²c)  4a²b²
=(cb²a²)² (2ab)²
=(cb^22*a*ba^2)*(cb^2+2*a*ba^2) 
Correction:
For the general sum from k=2 to k=n
Sum=((n1)!+(1)^n)/(n1)!