Prove by induction that... $1+3+5+7+...+(2n+1)=(n+1)^2$ for every $n \in \mathbb N$

Question

I'm not too sure exactly how to approach this question. Would anyone be able to give me any helpful advice or some sort of direction? I have a little problem with induction.

Prove by induction that:

$$1 + 3 + 5 + 7 +...+ (2n+1) = (n+1)^2 \quad\forall n \in \mathbb N $$

Let this be true for $n=m$ For $n=m+1, \sum_{r=1}^{m+1}(2r+1)=(m+1)^2+2m+3=[(m+1)+1]^2$ Finally establish the base case i.e., for $n=1$ — lab bhattacharjee, Sep 21 '14 at 12:41
@Semsem. Don't worry, please ! If every single time I made something similar they gave me a cent, taking into account my age, I should be a millionaire (may be billionaire !) ! Cheers. — Claude Leibovici, Sep 21 '14 at 12:57
@Semsem Let it go, please. What Claude just wrote is certainly true for me, as well! I'd just put it this way: If every such mistake cost me a cent, I'd be broke! — amWhy, Sep 21 '14 at 13:00

score 1 · Answer 1 · answered Sep 21 '14 at 12:49

1

For $n=1$ we have that $1+3=(1+1)^2=4$. Suppose that the relation is true for $n=m$ then $$1+3+\cdots +(2m+1)=(m+1)^2$$ then $$1+3+\cdots +(2m+1)+(2(m+1)+1)\\=(m+1)^2+(2(m+1)+1)\\=((m+1)+1)^2=(m+2)^2$$

answered Sep 21 '14 at 12:49

Semsem

7,651

I considered $m+1$ as one term and use the form $a^2+2a+1=(a+1)^2$ – Semsem Sep 21 '14 at 12:53

Aaron Maroja · Answer 2 · 2014-09-21T13:07:39.083

You should show that if $X = \lbrace n\in \mathbb{N}; 1 + 3 + ... + (2n+1) = (n+1)^2\rbrace \subset \mathbb{N}$ is such that $0 \in X$ and if $n \in X \Rightarrow n+1 \in X$ then $X=\mathbb{N}$.

Now, for $n=0$ we have

$$1 = 1^2 = 1$$

Suppose that for $n=k$ the assertion is valid. We must then prove that it holds for $n=k+1$. In fact

$$\begin{align} 1 + 3 + ... + (2k+1) + (2(k+1) + 1) = & (k+1)^2 + 2(k+1) + 1\\ = & k^2 + 2k + 1 + 2k + 2 + 1 = k^2 + 4k + 4 = (k+2)^2\end{align}$$

Then $n=k+1 \in X$.

By induction we have that it is true for all $n \in \mathbb{N}$.

As an exercise of it try to show that

i. $n \geq 4 \Rightarrow n! > 2^n$

ii. $2(1 +2 + ...+ n) = n(n+1)$

iii. $(a-1)(1+a+...+a^n) = a^{n+1} -1$

score 1 · Accepted Answer · answered Sep 21 '14 at 12:53

Let $P(n)$ be the statement $1 + 3 + 5 + \cdots + (2n + 1) = (n + 1)^2$.

The first thing you need to do is to show that $P(1)$ is true. In this case that means showing that $$1 + (2 \cdot 1 + 1) = (1 + 1)^2$$

Now that you have established that the proposition $P(n)$ is true for some integer $m$ (since it is true for $n = 1$), you may assume that $P(m)$ holds for some $m \in \mathbb{N}$.

The final step is to show that $P(m) \Rightarrow P(m + 1)$. Since $P(1)$ holds, establishing that $P(m) \Rightarrow P(m + 1)$ demonstrates that $P(n)$ holds for each $n \in \mathbb{N}$ since $P(1) \Rightarrow P(2) \Rightarrow P(3) \Rightarrow \cdots$.

In this case, to prove $P(m + 1)$ holds, you must show that

$$1 + 3 + 5 + \cdots + (2m + 1) + [2(m + 1) + 1] = [(m + 1) + 1]^2$$

under the assumption that $P(m)$ holds, that is, under the assumption that

$$1 + 3 + 5 + \cdots + (2m + 1) = (m + 1)^2$$

Prove by induction that... $1+3+5+7+...+(2n+1)=(n+1)^2$ for every $n \in \mathbb N$

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