Posted by **Gerald** on Saturday, August 21, 2010 at 12:27pm.

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)

- MATH -
**MathMate**, Saturday, August 21, 2010 at 8:57pm
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/(k-1)!

for k=2 to n is

((n-1)!+(-1)^k)/(k-1)!

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+c²-2a²c-2b²c) - 4a²b²

=(c-b²-a²)² -(2ab)²

=(c-b^2-2*a*b-a^2)*(c-b^2+2*a*b-a^2)

- MATH -
**MathMate**, Saturday, August 21, 2010 at 9:03pm
Correction:

For the general sum from k=2 to k=n

Sum=((n-1)!+(-1)^n)/(n-1)!

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