Prove this Using Mathemetical Induction

Question

using mathematical induction, prove that 1+5+9+...+(4n+1)=(n+1)(2n+1)

 

Please explain this as thoroughly as you can! thank you. 

Answer 1

The basic principal of induction says....prove it that 1 works, assume it works for k and then prove that it works for k+1.

The reasoning behind induction is as follows. If you showed the things above, then if it works for 1, you proved that it works for 2 since 2 is 1 more than 1. And if it works for 2, then it works for 2+1 or 3....and if it works for 3, then it works for 3+1 or 4......and so on for all integers. You do have to show the first one works and then prove that 1 more than the previous term works so you can build this recursive situation.

 

Now to your particular situation:

1

∑      4n+1 = 1 + 5 = 6       using the formula you get (1+1)(2*1+1) = 2*3 = 6 so this works for n=1

n=0

 

Assume

k

∑       4n+1 = (k+1)(2k+1)

n=0

 

PROVE

k+1

∑          4n+1 = (k+1 +1)(2(k+1) +1) = (k+2)(2k+3)

n=0

 

So,

1+5+9+.....(4k+1) + [4(k+1) +1] is what the summation symbol of what we want to prove means. We can now take what we assumed and replace all but the last term with our assumption to get....

 

k

∑        4n+1         + [4(k+1)+1]     which is      (k+1)(2k+1)     + [4k+5]

n=0

 

That's 2k²+3k+1   +4k+5     or

 

2k²+7k+6 which factors into

(k+2)(2k+3).

 

When you plug k+1 into the formula you're trying to prove you get (k+1+1)(2(k+1)+1) = the same thing.

So you're done.

 

Hope this makes some sense to you.