Problems

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 $$1+3+5+\cdots +(2n-1)=n^2.$$

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) $$1^2+2^2+3^2+\cdots +n^2=\frac{1}{6}n(n+1)(2n+1)$$

(b) $$1^2+3^2+5^2+\cdots +(2n-1{)}^2=\frac{1}{3}n(2n-1)(2n+1).$$

Using mathematical induction prove that $2^n>n$ for all natural numbers.

Let $n$ be some positive number. It is obvious that $$2n-1<2n.$$ Take another positive number $a$, and multiply both sides of the inequality by $(-a)$ $$-2na+a<-2na.$$ Now, subtracting $(-2na)$ from both sides of the inequality we get $a<0$.

Thus, ALL POSITIVE NUMBERS ARE NEGATIVE!

Suppose $a\ne b$. We can write $$-a=b-(a+b)$$ and $$-b=a-(a+b)$$ Since $(-a)b=a(-b)$, then $$(b-(a+b))b=a(a-(a+b))$$ Removing the brackets, we have $$b^2-(a+b)b=a^2-a(a+b)$$ Adding ${\left(\frac{a+b}{2}\right)}^2$ to each member of the equality we may complete the square of the differences of two numbers $${(b-\frac{a+b}{2})}^2={(a-\frac{a+b}{2})}^2$$ From the equality of the squares we conclude the equality of the bases $$b-\frac{a+b}{2}=a-\frac{a+b}{2}.$$ Adding $\frac{a+b}{2}$ to both sides of equality we get $a=b$. Therefore, WE HAVE SHOWED THAT FROM $a\ne b$ IT FOLLOWS $a=b$.

Let $x$ be equal to 1. Then we can write $x^2=1$, or putting it differently $x^2-1=0$. By using the difference of two squares formula we get $$(x+1)(x-1)=0$$ Dividing both sides of the equality by $x-1$ we obtain $$x+1=0,$$ in other words $x=-1$. But earlier we assumed $x=1$. THUS $$-1=1!$$

In every right-angled triangle the arm is greater than the hypotenuse. Consider a triangle $ABC$ with right angle at $C$.

The difference of the squares of the hypothenuse and one of the arms is $A{B}^2-B{C}^2$. This expression can be represented in the form of a product $$A{B}^2-B{C}^2=(AB-BC)(AB+BC)$$ or $$A{B}^2-B{C}^2=-(BC-AB)(AB+BC)$$ Dividing the right hand sides by the product $-(AB-BC)(AB+BC)$, we obtain the proportion $$\frac{AB+BC}{-(AB+BC)}=\frac{BC-AB}{AB-BC}.$$ Since the positive quantity is greater than the negative one we have $AB+BC>-(AB+BC)$. But then also $BC-AB>AB-BC$, and therefore $2BC>2AB$, or $BC>AB$, i.e. THE ARM IS GREATER THAN THE HYPOTENUSE!