Exercise 8.4.1

Question

For $n \in N$, let
$$N\# = n + (n-1) + (n-1) + \cdots + 2 + 1$$
\begin{enumerate}[label=(\alph*)] 
\item Without looking ahead, decide if there is a natural way to define $0\#$. How about $(-2)\#$? Conjecture a reasonable value for $\frac{7}{2}\#$.
\item Now prove $n\# = \frac{1}{2}n (n+1)$ for all $n \in \mathbf{N}$, and revisit part (a).
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item Noting that $N\# = (N-1)\# + N$ and $1\# = 1$, we could have $0\# = 0$, $-1\# = 0$, and $-2\# = 1$. $7/2\#$ could just be defined to be the result from linearly interpolating between $3\# = 6$ and $4\# = 10$ to get $8$.
\item This is obviously true for $n = 1$, and
$$(n+1)\# = n+1+n\# = n+1 + \frac{1}{2}n(n+1) = (n+1)\left(1 + \frac{n}{2}\right) = (n+1)\left(\frac{n+2}{2}\right)$$
which proves the formula by induction.
 \end{enumerate}

Exercise 8.4.1

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Exercise 8.4.1

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