integer values

Question

"Show that (n+3)^2-(n-3)^2 is an even number for all positive integer values of n"

 

I would like to know how to workout this type of question.

 

Thank you in advance.

Answer 1

Hi, let us proceed as follows:

 

(n+3)^2 = n^2 + 2(n)(3) + 9         How you may ask? This is, because, (a+b)^2 = a^2 + 2 ab + b^2

 

(n-3)^2 = n^2 - 2(n)(3) + 9                                           "   "       "        (a-b)^2 = a^2 - 2ab + b^2

 

So, what is (n+3)^2 - (n-3)^2 ?  (what the question says)

 

From above, it is:   (n^2 + 6n +9) - (n^2-6n +9)

 

Simplyfying,  it is:  n^2 + 6n + 9 - n^2 + 6n -9

 

              which is,  =  12n      (note that n^2 & -n^2 cancel out; also, +9 & -9 cancel out) 

 

       Our answer is  12n 

 

Now, try any positive integer for 'n'.  12n will always be an even number.

 

Example:  12 X 3 = 36, which is, even

 

                12 X 8  = 96, an even number

 

                 12 X 13 = 156, even

 

                 12 X 17 = 204, even

 

                 etc., etc.

 

           So, for all positive integers, odd or even, the answer is an even number.