How do I prove this equation by mathematical induction?

Question

I am new to induction and am stuck on the following formula. I have completed the base induction step, but I don't even really know where to start when proving $P(k+1)$. Any help would be greatly appreciated. Thanks! (P.S. this is for all integers $n \geq 0$).

$P(n)$: $$\sum_{j=0}^n (4j+1) = 2n^2+3n+1$$

$P(k+1)$: $$\sum_{j=0}^{k+1} (4j+1) = 2(k+1)^2+3(k+1)+1$$

Answer 1

You have $\sum_{j=0}^k(4j+1) = 2k^2 + 3k + 1$, so when you take $\sum_{j=0}^{k+1}(4j+1)$ you are just adding one extra term: $4(k+1)+1$. Add this to your previous total and see if you can factor the resulting polynomial into the desired form.

Answer 2

Hint:

Rewrite the inductive hypothesis in function of $k+1$: $$2k^2+3k+1=2(k+1)^2-(k+1).$$

Answer 3

Try this: $$\sum_{j=0}^n (4j+1) = 2n^2+3n+1$$

First, note that in the n+1 case the difference on the Left is 4(n+1)+1=4n+5. Secondly, the right hand side has difference 4n+2+3= 4n+5 .These are equal, so they change by the same amount, when the variable changes by 1. Therefore, they are equal for any natural number n, greater than a proven base case.