## Consider an equilateral triangle in which straight lines are drawn from each vertex to the middle of the opposite side, such as in the size $1$ triangle in the sketch below.

Sixteen triangles of either different shape or size or orientation or location can now be observed in that triangle. Using size $1$ triangles as building blocks, larger triangles can be formed, such as the size $2$ triangle in the above sketch. One-hundred and four triangles of either different shape or size or orientation or location can now be observed in that size $2$ triangle.

It can be observed that the size $2$ triangle contains $4$ size $1$ triangle building blocks. A size $3$ triangle would contain $9$ size $1$ triangle building blocks and a size $n$ triangle would thus contain $n^2$ size $1$ triangle building blocks.

If we denote $T(n)$ as the number of triangles present in a triangle of size $n$, then

\begin{align}

T(1) &= 16\\

T(2) &= 104

\end{align}

Find $T(36)$.

### The function $T(n)$, which represents the number of unique triangles in size $n$ triangle, gives the counts:

\begin{align*}

T(1) &= 16, \\

T(2) &= 104.

\end{align*}

It implies that the number of triangles increases while $n$ increases. It’s would be a good idea to compute $T(3)$ to grasp the relationship between $n$ and $T(n)$.

For $n=1$, the increase in the number of triangles from $0$ is $16 = 16$.

For $n=2$, the increase in the number of triangles from $n=1$ is $104-16 = 88$.

Assume that for $n=3$, the increase in the number of triangles from $n=2$ is twice that of $n=2$ to $n=1$. It gives:

$T(3) = 104 + 2*88 = 280$.

Consider that the increase in the number of triangles from $n$ to $(n+1)$ is twice that of $(n-1)$ to $n$.

Therefore, for $n=36$, the number of triangles could be calculated using the formula:

\begin{align*}

T(n) &= T(n-1) + 2^{(n-1)}*88 \\

&= T(n-2) + 2^{(n-2)}*88 + 2^(n-1)*88 \\

&= T(1) + 2^0 * 88 + 2^1 * 88 + \cdots + 2^{(n-1)}*88.

\end{align*}

It is a geometric series. Use the formula for the sum of a geometric series:

$$S = a * (1 – r^n) / (1 – r),$$

where:

– $S$ is the sum of the geometric series,

– $a$ is the first term,

– $n$ is the quantity of items,

– $r$ is the common ratio between the items.

Substite the variables:

\begin{align*}

T(36) &= 16 + 88 * (1 – 2^{35}) / (1 – 2) \\

&= 16 – 88 * (1 – 2^{35}) \\

&= 16 – 88 + 88 * 2^{35} \\

&= 2^{37} * 11 – 72 \\

&= 302875106592252.

\end{align*}

Therefore, $T(36) = 302875106592252$.

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