Proof by induction

Question

1 × 3 + 3 × 5 + 5 × 7 + . . . + (2n − 1)(2n + 1) = n(4n² + 6n − 1)/3.

 

 

 

Answer 1

The proof structure follows what it was in the last question. Here, as Kenneth S. says, except with variables matching my last answer rather than his answer above (my n+1 is his n, my n is his k), show that

 

(n+1)(4(n+1)²+6(n+1)-1)/3

= n(4n² + 6n - 1)/3 + (2n+1)(2n+3)

 

Hint: use a common denominator of 3, and multiply everything out on both sides.

 

On the left is what is to be shown equal to what is on the right: the inductive hypothesis plus what is added to the sum in the inductive step.

Answer 2

All proofs by mathematical induction follow the same steps. Algebra skills are required.

 

1. Test if proposition is true for n=1.  1 times 3 does equal 1(4+6-1)/3.

2. Assume true for n=k.

3. Show that it follows, algebraically, that it must then be true for n = k+1, based on writing the statement to be proved, with n=k on each side, AND THEN ADDING the next term to both sides.

 

That means add (2k+1)(2k+3) to each side. Then YOU must factor the right side and show that this equation's two sides correspond to "proposition k+1" and thus, proposition k (assumed) implies the truth of proposition k+1. (Then this implies,by repeated applications, truth in general.)

 

So get going on the necessary algebra!