Proving by Mathematical Induction

[Phyzwizz]

so the problem is 3+7+11+15+...+(4n-1) = n(2n+1)

so I know that step 1 is to plug in 1 for the right side and check that it equals three...
3=1(2(1)+1) and yes it equals 3

Then I know that you assume that 3+7+11+15+...+(4n-1) = n(2n+1)

The next step is where I get confused I know that you can change 3+7+11+15 into n(2n+1) so that on the left side you get n(2n+1)+(4n-1), but I know that you have to do something with the right side in this step as well please help me figure this out, I'm sure its something really easy that I'll feel stupid for not remembering but help would be great.

 

[SammyS]

so the problem is 3+7+11+15+...+(4n-1) = n(2n+1)

so I know that step 1 is to plug in 1 for the right side and check that it equals three...
3=1(2(1)+1) and yes it equals 3

Then I know that you assume that 3+7+11+15+...+(4n-1) = n(2n+1)
...

Then assuming that, you must show that it's true for n+1.

In other words:

Assume that 3+7+11+15+...+(4n-1) = n(2n+1)

Show that this leads to 3+7+11+15+...+(4n-1)+(4(n+1)-1) = (n+1)(2(n+1)+1)​

 

[HallsofIvy]

If, for a particular n, 3+7+11+15+...+(4n-1) = n(2n+1), then
3+7+11+15+...+(4n-1)+ [4(n+ 1)- 1] = n(2n+1)+ 4n+ 3= 2n^2+ 5n+ 3
See how you can factor that.