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I am having trouble in identifying how $n=k+1$ differs from $n=k$.

Assume it is true for some integer $k \ge 1$, that is $1+3+(2k-1)+(2k+1)+(2k-1)+3+1$. I need help getting to how $n=k$ implies $n=k+1$ in this case.

Adam6
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    Suggested reading: How To Write a Clear Induction Proof. – JMoravitz May 05 '20 at 01:57
  • $\color{blue}1+\color{brown}3+\cdots+\color{green}{(2n-1)}+(2n+1)+\color{blue}{(2n-1)}+\color{brown}{(2n-3)}+\cdots+\color{green}1 $ $=\color{blue}{1+(2n-1)}+\color{green}{3+(2n-3)}+...+\color{brown}{(2n-1)+1}+(2n+1)$ $=2n\times n+(2n+1)=n^2+(n+1)^2$ – J. W. Tanner May 05 '20 at 02:05
  • Prove $1 + 3 + ...... (2n-1)= n^2$ first. Then that wil mean $1 + 3 + ...... +(2n-1) + (2n+1)$ because $2n+1 = 2(n+1) -1$. Then $(2n-1) + .... + 3 + 1 = n^2$ and we are done.... $1 + 3 + ...... +(2n-1) + (2n+1) + (2n-1) + .... + 3 + 1=(n+1)^2 +n^2$ – fleablood May 05 '20 at 02:08

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So, finish your base case and set up your induction hypothesis.

We suppose that it is true for some integer $k\geq 1$ that we have $1+3+\color{red}{\dots}+(2k-1)+(2k+1)+(2k-1)+\color{red}{\dots}+3+1=k^2+(k+1)^2$

(you forgot your dots there...)

We try to use this hypothesis to prove the case for when $n=k+1$. Now, it follows that:

$$\begin{array}{l}1+3+\dots+(2k-1)+(2k+1)+\color{blue}{(2k+3)+(2k+1)}+(2k-1)+\dots+3+1\\=\dots\\=\dots~~\text{by induction hypothesis}\\=\dots\\\vdots\\=(k+1)^2+(k+2)^2\end{array}$$

Note that anywhere where $k$ appeared previously, $k+1$ now appears. I went ahead and simplified some of those... but where we had $k^2$ we now have $(k+1)^2$. Where we had $(k+1)^2$ we now have $((k+1)+1)^2$ or in other words $(k+2)^2$. Where the largest term in the middle of the summation was previously $(2k+1)$ the largest term in the middle of the summation was now $(2(k+1)+1)$ or simplified $(2k+3)$.

Now... as for a hint on actually applying your induction hypothesis, maybe take the middle terms that I colored blue and add it later. If you were to add only the black terms in the above, the induction hypothesis tells you how you can simplify that.

JMoravitz
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  • Thanks, I have the base case on my paper, I just didn't write it out. – Adam6 May 05 '20 at 02:08
  • I definitely got it now. I totally forgot to substitute the induction hypothesis in for $n=k$. Very simple after realizing that, so thanks for the help. My brain is fried from studying for finals. – Adam6 May 05 '20 at 02:20
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From here $$1 + 3 + \ldots + (2k - 1) + (2k + 1) + (2k - 1) + \ldots + 3 + 1$$ to here $$1 + 3 + \ldots + (2k - 1) + (2k + 1) + \underbrace{(2k + 3) + (2k + 1)}_\text{These are new} + (2k - 1) + \ldots + 3 + 1.$$ Let me know if you need more help.

user780985
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