Question

If we multiply three consecutive integers and then add the middle number to the result, the answer is always the middle number cubed.

5 × 6 × 7 + 6 = 216 = 63

10 × 11 × 12 + 11 = 1331 = 113

15 × 16 × 17 + 16 = 4096 = 163

Use what you have learned in this lesson to write a third-degree polynomial expression in factored form and standard form that represents this pattern. Show that the factored form is equivalent to the expression written in standard form. Substitute three different values for the variable to prove this pattern works for other numbers.

Answer 1

Step-by-step explanation:

Let use have a interger,n

Thus three consectice intergers can be represented by

$n$

$n + 1$

$n + 2$

Multiply, and we get

$n(n + 1)(n + 2)$

$n( {n}^{2} + 3n + 2)$

$n {}^{3} + 3 {n}^{2} + 2n$

Then add the middle number, n+1.

${n}^{3} + 3n {}^{2} + 3n + 1$

We must prove that the middle number cubed = the equation.

Using the binomial theroem,

$(n + 1) {}^{3} = {n}^{3} + 3( {n}^{2} 1 {}^{1} ) + 3n( {1}^{2} ) + 1 {}^{3}$

$(n + 1 ){}^{3} = {n}^{3} + 3 {n}^{2} + 3n + 1$

So this is indeed true.

You can do the subsitue variables for n to prove it.