Finding how many unit triangles in 10 rows? Okay, so picture a triangular arrangement that consists of a number of unit triangles (side lengths that equal to 1). The series for the total number of unit triangles in the "diagram" would be 1 + 3 + 5 + 7.
I need to find how many unit triangles are there if there are 10 rows in the triangular arrangement.
I used this formula: $S_{n}$ = $\frac {n}{2} [2t_{1} + (n - 1)d]$ with my:


*

*$n$ = 10

*$t_1$ = 1

*$d$ = 2


here's my work:
$S_{10}$ = $\frac {10}{2} [2(1) + (10 - 1)2]$
$S_{10}$ = 5 [2 + 38]
$S_{10}$ = $200$
The answer in the textbook is $100$,what did I do wrong?
 A: You made a type with your formula. The complete calculation should look like this.
$$S_{10}=[2(1)+(10-1)2]$$
$$S_{10}=5[2+18]$$
$$S_{10}=100$$
You made a simple typo when you were working on this expression:$(10-1)2$
A: I did this mentally, without recourse to any diagram (so I hope it fits). It's based on what you provided as an example sequence.
For $10$ rows, the series would be $1 + 3 + 5 + ... + T_{10}$
where $T_{10}$ would be the $10$th odd number (not really necessary to bother computing it right now, even if it's easy.
Let the sum of the above be $S$.
Now, let's make it a series that we're all very familiar with.
Add $1$ to each term to get:
$2 + 4 + 6 + ... + (T_{10} + 1)$
Now halve each term to get:
$1 + 2 + 3 + ... + \frac{(T_{10}+ 1)}2$
You should now be able to recognise that the last term is just $10$ since you should have ten terms and you're just summing the first ten natural numbers.
So:
$\frac{S+10}{2} = 1 + 2 + 3 + ... + 10 = \frac 12 (10)(10+11)$
(Here I used the well known sum for a natural number series, which is the same as the sum of the first ten triangular numbers. You can also easily derive this from an arithmetic series formula).
$S+10 = (10)(11)$
$S = (10)(10) = 100$
As required.
