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Prove that the $n$-th triangular number is:

$$T_n = \dfrac{n(n+1)}{2}$$

I did this:

Base case: $\frac{1(1+1)}{2}=1$, which is true.

Then I assumed that $T_k=\frac{k(k+1)}{2}$ is true.

$$T_{k+1} = \frac{(k+1)(k+1+1)}{2}$$

I'm not sure what to do next. What is this supposed to be equal to?

MrMazgari
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Segmentation fault
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2 Answers2

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You have to think about the nature of the triangular numbers: the $n$-th triangular number is the number of dots created by $n$ layers of dots stacked upon each other: the first (top) layer has $1$ dot, the next (below it) has $2$ dots, etc. The $n$-th and last layer of the $n$-the triangular number has $n$ dots:

enter image description here

Now, your inductive hypothesis is that the $k$-the triangular number consists of $\frac{k(k+1)}{2}$ dots, i.e. that

$$T_k = \frac{k(k+1)}{2}$$

Using that hypothesis, you have to show that the $k+1$-th triangular number has $\frac{(k+1)(k+2)}{2}$ dots. But note: the $k+1$-th triangular number adds a layer of $k+1$ dots to the $k$-th triangular number. That is, we know that:

$$T_{k+1}=T_k +(k+1)$$

So, use that fact, together with the inductive hypothesis, to show what you need to show, i.e. that

$$T_{k+1}=\frac{(k+1)(k+2)}{2}$$

Bram28
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    The gist to induction is basically, write sequence for n = T_n, then sequence for n + (n+1) = T_n+1, then replace the sequence for n but with T_n, simplify, etc etc, then prove they are equal. Right? – Segmentation fault Jan 03 '20 at 01:23
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    @SilenceOnTheWire Correct! :) – Bram28 Jan 03 '20 at 01:29
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For the inductive step, we have:

\begin{align} T_{k+1}&=T_k+(k+1) \\ &=\frac{k(k+1)}{2} + (k+1) \\ &=\frac{(k+1)(k+2)}{2} \end{align}

This is the desired result. Hence, true by the principle of mathematical induction...

MrMazgari
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