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I'm reading the following book: http://www.cs.princeton.edu/courses/archive/spring10/cos433/mathcs.pdf.

In page 26, they attempt to prove the following theorem using Induction:

For all $n \in \mathbb{N}$: $$1 + 2 + \cdots + n = \frac{n(n+1)}{2}$$

In order to prove the theorem we need to prove the following statements:

  • $P(0)$ is true.
  • For all $n \in \mathbb{N}$, $P(n)$ implies $P(n + 1)$.

Proving the first one is easy:

$$P(0) = 0 = 0 * (0 + 1) / 2$$ $$= 0 = 0 * 1 / 2$$ $$= 0 = 0 / 2$$ $$= 0 = 0$$

In order to prove the second, they state the following:

$$1+2+\cdots + n + (n+1) = \frac{n(n+1)}{2} + (n+1) \tag{1}$$ $$ = \frac{(n+1)(n+2)}{2} \tag{2}$$

I don't understand how they get from (1) to (2) and how that proves that proves that for all $n \in \mathbb{N}$, $P(n)$ implies $P(n + 1)$ is true.

I'm clearly missing something in the process. Can someone clear the fog for me?

jviotti
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2 Answers2

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Just do the math.

$$\frac{n(n+1)}{2} + (n+1) = \frac{n(n+1)+2(n+1)}{2} = \frac{(n+2)(n+1)}{2}$$

For your second problem, note that you proved that $$1+2+\ldots+n+(n+1) = \frac{(n+2)(n+1)}{2}.$$

The initial statement is $1+2+\ldots+k = \frac{k(k+1)}{2}$, with another letter. If you do $k=n+1$ you will get $1+2+\ldots+(n+1) = \frac{(n+1)((n+1)+1)}{2} = \frac{(n+2)(n+1)}{2}$, and that is what you just proved.

Integral
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\begin{align*} \frac{n\left(n+1\right)}{2}+\left(n+1\right) & =\frac{n\left(n+1\right)}{2}+\frac{2\left(n+1\right)}{2}\\ & =\frac{n^{2}+n+2n+2}{2}\\ & =\frac{n^{2}+3n+2}{2}\\ & =\frac{\left(n+2\right)\left(n+1\right)}{2} \end{align*}

parsiad
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