Question

1. Prove that the product of the second and third numbers out of four consecutive whole numbers is two greater than the product of the first and the fourth numbers.

2. Prove that the square of the second number out of three consecutive odd numbers is four greater than the product of the first and third numbers.

Answer 1

1.

(x+1)(x+2)>x(x+3)

x^2+2x+x+2>x^2+3x

x^2+3x+2>x+3x

2>0

2.

(x+2)^2>x(x+4)

x^2+4x+4>x^2+4x

4>0

Explanation:

1.

Suppose x is the lowest number in the set of 4 consecutive numbers, then x+1, x+2, and x+3 are then the next numbers in the sequence since they are the next whole numbers.

Then, to follow the conditions of the problem, we make the equation:

(x+1)(x+2)>x(x+3)

Using the formula of (a+b)(c+d)=ac+ad+bc+bd, we can expand both sides of the equation into this:

x^2+2x+x+2>x^2+3x

=x^2+3x+2>x^2+3x

We can then cancel x^2+2x from both sides and we are left with 2>0.

2.

Again, suppose x is the lowest value in the set of consecutive numbers. In this case, odd numbers increase by 2 each time so the series of numbers are:

x, x+2, x+4

Therefore, to follow the conditions of the problem, we make the equation:

(x+2)^2>x(x+4)=

(x+2)(x+2)>x(x+4)

Using the same formula of (a+b)(c+d)=ac+ad+bc+bd, we get:

x^2+4x+4>x^2+4x

Cancel x^2+4x from both sides and we are left with 4>0

Hope this helps!