\begin{tabular}{ccccc}
& 0 & 00 & 000 & 0000 \\
1 & 3 & 6 & 10 & 15
\end{tabular}

When [tex]$n=1$[/tex], there is 1 dot. When [tex]$n=2$[/tex], there are 3 dots. When [tex]$n=3$[/tex], there are 6 dots. Notice that the total number of dots [tex]$d(n)$[/tex] increases by [tex]$n$[/tex] each time.
Use induction to prove that [tex]$d(n)=\frac{n(n+1)}{2}$[/tex].

Part A

Prove the statement is true for [tex]$n=1$[/tex].

Type your answer in the box.



Answer :

To prove by induction that [tex]\( d(n) = \frac{n(n+1)}{2} \)[/tex], we first need to confirm that the statement is true for [tex]\( n = 1 \)[/tex].

### Base Case:
Let's start with [tex]\( n = 1 \)[/tex]:

When [tex]\( n = 1 \)[/tex], there is 1 dot. According to the formula [tex]\( d(n) = \frac{n(n+1)}{2} \)[/tex]:

[tex]\[ d(1) = \frac{1(1+1)}{2} = \frac{1 \times 2}{2} = \frac{2}{2} = 1 \][/tex]

We observe that the number of dots when [tex]\( n = 1 \)[/tex] is indeed 1, which matches our derived value using the formula. Therefore, the statement holds true for [tex]\( n = 1 \)[/tex].

Hence, the statement [tex]\( d(n) = \frac{n(n+1)}{2} \)[/tex] is true for the base case [tex]\( n = 1 \)[/tex].

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