Proof by induction (summation)

Question

I need to prove that

$$\sum_{k=0}^n (2k+1) = (n+1)^2$$

What I tried:

$$\sum_{k=0}^{n+1} (2k+1) = \sum_{k=0}^n (2k+1) + (n+1) = (n+1)^2 + (n+1)$$ $$(n+1)[(n+1)+1] = (n+1)(n+2)$$

Which is not equal to $((n+1)+1)^2 = (n+2)^2$

Any tip? Thanks in advance.

Now if I replace by the Inductive Hypotesis I got $(n+1)^2 + (2k + 3)$. No idea how to continue as I now have 2 variables. — David Zavalla, Apr 23 '17 at 00:39
Alright so I have $n^2 + 4n + 4$. Then $(n+2)^2 = n^2+2n2+2^2 = n^2+4n+4$ so I'm done right? — David Zavalla, Apr 23 '17 at 00:55
Alternative hint: prove (by induction or otherwise) that $\sum_{k=0}^n k = \frac{1}{2}n(n+1),$. Then:
$$\sum_{k=0}^n (2k+1) = 2 \sum_{k=0}^n k + \sum_{k=0}^n 1 = n(n+1) + n + 1= (n+1)^2$$ — dxiv, Apr 23 '17 at 01:24

score 0 · Accepted Answer · answered Apr 23 '17 at 01:21

Base case: at $n = 1$ we have $2(0) + 1 + 2(1) + 1 = 4 = (1+1)^2$.

Assume that the relation works up to $n$, then $$\sum_{k=0}^{n+1} (2k+1) = \sum_{k=0}^{n} (2k+1) + (2(n+1)+1) = (n+1)^2 + 2n + 3 = n^2 + 4n + 4 = (n+2)^2 = ((n+1)+1)^2$$ This completes the proof.

Proof by induction (summation)

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