There are \(n\) lines on a plane, and all the lines intersect at exactly one point. Prove that the lines divide the plane into \(2n\) parts.
There are \(n\) lines on a plane, no two lines are parallel, and no three lines cross at one point. Show that those lines dived the plane into \(\frac{n(n+1)}{2}+1\) regions.
In a sequence 2, 6, 12, 20, 30, ... find the number
(a) in the 6th place
(b) in the 2016th place.
Using mathematical induction prove that \[1 +3 +5 +\dots + (2n-1) = n^2.\]
Circles and lines are drawn on the plane. They divide the plane into non-intersecting regions, see the picture below.
Show that it is possible to colour the regions with two colours in such a way that no two regions sharing some length of border are the same colour.
Numbers \(1,2,\dots,n\) are written on a whiteboard. In one go Louise is allowed to wipe out any two numbers \(a\) and \(b\), and write their sum \(a+b\) instead. Louise enjoys erasing the numbers, and continues the procedure until only one number is left on the whiteboard.
What number is it? What if instead of \(a+b\) she writes \(a+b-1\)?
Prove that
(a) \[1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{1}{6} n (n+1)(2n+1)\]
(b) \[1^2 + 3^2 + 5^2 + \dots + (2n-1)^2 = \frac{1}{3} n (2n-1)(2n+1).\]
Using mathematical induction prove that \(2^n>n\) for all natural numbers.
Take any two non-equal numbers \(a\) and \(b\), then we can write \[a^2 - 2ab + b^2 = b^2 - 2ab + a^2\] Using the formula \((x-y)^2 = x^2 - 2xy + y^2\), we complete the squares and rewrite the equality above as \[(a-b)^2 = (b-a)^2.\] As we take a square root from the both sides of the equality, we get \[a-b = b-a.\] Finally, adding to both sides \(a+b\) we get \[\begin{aligned} a-b + (a+b) &= b-a + (a+ b)\\ 2a&= 2b\\ a&=b. \end{aligned}\] Therefore, All NON-EQUAL NUMBERS ARE EQUAL! (This is gibberish, isn’t it?)
Consider equation \[x-a=0\] Dividing both sides of this equation by \(x-a\), we get \[\frac{x-a}{x-a} = \frac{0}{x-a}.\] But \(\frac{x-a}{x-a}=1\) and \(\frac{0}{x-a}=0\). Therefore, we get \[1=0.\]