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I just started working on Induction, and I have one particular problem that I don't understand:

Prove that $1+3+5+...+(2n−1)=n^2$ for any integer $n≥1.$

  • $n = 1$ :

    $1 = 1^2$

  • $n = k$ :

    $1+3+5+...+(2k−1)=k^2$

  • $n = k+1$ : (this is where I have a problem)

    I thought that you simply substitute k+1 for n, yielding $1+3+5+...+(2k+1)$, but the correct equation is actually $1+3+5+...+(2k−1)+(2k+1)$. Where did that extra term $2k-1$ come from?

My apologies if this is a simple answer, but this is all new to me. Thanks in advance

user3170251
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1 Answers1

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Don't get confused. For example, if I write the expression $2+4+6+\cdots +10$, isn't it the same as $2+4+6+\cdots 8+10$? (the sum of even integers from $2$ to $10$, inclusive). They're just including the second last term, that's all. This will be needed in order to apply the induction hypothesis (from the case $n=k$, you need the sum $1+3+5+\cdots +(2k-1)$ to appear somewhere).

This is a very common idea when trying to prove identities involving sums (or products) of $n$ terms by induction on $n$: Isolate the $n+1$ term and notice that the remaining part will be the induction hypothesis.

Daniel
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