Sigma notation using induction

Question

Suppose that n∈N, $$\sum_{k=1}^n (2k+1) = n^2+2n$$

Base Case:n=1

⟹2∗1+1=3=12+2∗1

the base case holds true

I.H, Assume its true for $$\sum_{k=1}^{n} (2k+1) = n^2+2n$$

Then;

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

$$\implies\sum_{k=1}^{n+1} (2k+1) = (n+1)^2+2(n+1)$$ Im confused how to proceed next?

try splitting the sum in such a way that you can use the induction hypothesis you need to show that $$\sum_{k=1}^{n+1} (2k+1) = (n+1)^2+2(n+1)$$ — asddf, Jul 14 '17 at 23:29
$$\sum_{k=1}^{n+1} (2k+1) $$=$$\sum_{k=1}^{n} (2k+1) +2(n+1)+1$$ — asddf, Jul 14 '17 at 23:32
@asddf can u check my updated answer, is that what u wanted, splitting it? — TheGamer, Jul 14 '17 at 23:33
I saw your edit, the thing is you dont know whether $$\sum_{k=1}^{n+1} (2k+1) = (n+1)^2+2(n+1)$$ is true, you only know that the induction hypothesis is true, try using induction hypothesis on the 4th comment in the first sum which goes up to n and then $(n+1)^2 +2(n+1)$ should come out — asddf, Jul 14 '17 at 23:37
@TheGamer How is this not a duplicate of your other question: Proving $1+3+\cdots+(2n-1) = n^2$ using induction. Just because the index of summation is shifted by $1$ doesn't make it a different problem. — dxiv, Jul 14 '17 at 23:46

score 0 · Accepted Answer · answered Jul 14 '17 at 23:40

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

and let $n\ge 1$ be such that $$S_n=n^2+2n .$$

we must prove that $$S_{n+1}= (n+1)^2+2(n+1)$$

we have by definition $$S_{n+1}=\sum_{k=1}^{n+1}(2k+1)$$ $$=S_n+2 (n+1)+1$$ $$=n^2+2n+2 (n+1 )+1$$ $$=n^2+2n+1+2(n+1 )$$ $$=(n+1)^2+2 (n+1) $$ Qed.

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