Developing Algebraic Equations (number of squares in "staircase")

[Eew3456]

Trying to figure out what algebraic equation represents the total number of squares for the figures. I figured out that the height equation is n+1. The total number of squares isn’t the same as the typical staircase problem since figure one is the 3 squares.

 

[Otis]

what algebraic equation represents the total number of squares for the figures

Hi. If we count squares by adding row totals, does that give you an idea? (For completeness, I'll include an extra figure that has only one square.)

1 = 1
1 + 2 = 3
1 + 2 + 3 = 6
1 + 2 + 3 + 4 = 10
1 + 2 + 3 + 4 + 5 = 15

Have you seen sigma notation or any summation formulas – listed under 'Useful Shortcuts' here? (There are various ways to not count that extra square above.)

If not, then please tell us what your class is currently studying.
[imath]\;[/imath]

 

[Dr.Peterson]

View attachment 35096
Trying to figure out what algebraic equation represents the total number of squares for the figures. I figured out that the height equation is n+1. The total number of squares isn’t the same as the typical staircase problem since figure one is the 3 squares.

What is the "typical staircase problem", as you understand it? Maybe we can modify its answer to fit this. (Algebra is good at doing that.) Or maybe we can modify its method of solution.

 

[MarkFL]

What you are trying to find is closed form for the nth triangular number. It puts me in mind of a story, true or not, of a young Karl Gauss.

 

[MarkFL]

The nth triangular number is:

\(\displaystyle T_n=\sum_{k=1}^{n}(k)\)

Perhaps the simplest way to find this closed form is to write:

\(\displaystyle T_n=1+2+3+\dots+(n-2)+(n-1)+n\)

\(\displaystyle T_n=n+(n-1)+(n-2)\dots+3+2+1\)

Adding, we find we have n summands of n + 1:

\(\displaystyle 2T_n=n(n+1)\)

Thus:

\(\displaystyle T_n=\frac{n(n+1)}{2}\)

 

[Dr.Peterson]

View attachment 35096
Trying to figure out what algebraic equation represents the total number of squares for the figures. I figured out that the height equation is n+1. The total number of squares isn’t the same as the typical staircase problem since figure one is the 3 squares.

If I assume that the "staircase problem" you mention is the triangular numbers, as @MarkFL has presented it, you seem to be troubled only by the fact that this problem starts with 2 layers for n=1:

(That assumes your n is the number of the figure.)

If so, then your F(n) is just T_n+1. For example, F(1) = T₂ = 2(2+1)/2 = 3.

What formula does that give you, if you replace n with n+1?

 

[Tollens]

This looks like a finite math kind of question. I would look at the height.

n = 1 -> blocks = 1
n = 2 -> blocks = 3
n = 3 -> blocks = 6
n = 4 -> blocks = 10

So if you sum up the n numbers, you get the total blocks.

The general formula to find the number of blocks is n!

So for example if the height is 4 then n=4, and n! = (4+3+2+1) = 10

 

[MarkFL]

The factorial is a product, not a sum.

\(\displaystyle n!=\prod_{k=1}^{n}(k)\)

 

[Tollens]

The factorial is a product, not a sum.

\(\displaystyle n!=\prod_{k=1}^{n}(k)\)

Oh yeah, it is the E symbol, but I don't know how to put that in

 

[blamocur]

Oh yeah, it is the E symbol, but I don't know how to put that in

\sum : [imath]\sum[/imath] ? I believe the Greeks call it "sigma" ([imath]\Sigma[/imath])