, 2015

Diagram 1 illustrates that $T(3)=9$.

To determine $T(4)$, add 1 line segment to Diagram 1 as shown in Diagram 2.

We are told that this new (4$^{th}$) line segment must intersect each of the existing 3 line segments exactly once, creating 3 new points of intersection (labelled $1,2,3$).

This 4$^{th}$ line segment also adds 2 new endpoints (labelled $4$ and $5$) distinct from the previous 3 new points.

In addition, each of the points which exist in the illustration of $T(3)$ (Diagram 1) continue to exist in the illustration of $T(4)$ (Diagram 2) and are distinct from each of the new points which were added.

Therefore, we get $$\begin{aligned} T(4)& =T(3)+3+2\\ & =9+3+2\\ & =14\end{aligned}$$ Diagram 2 illustrates that $T(4)=14$.

To determine $T(5)$, add 1 line segment to Diagram 2 as shown in Diagram 3.

We are told that this new (5$^{th}$) line segment must intersect each of the existing 4 line segments exactly once, creating 4 new points of intersection (labelled $1,2,3,4$).

This 5$^{th}$ line segment also adds 2 new endpoints (labelled $5$ and $6$) distinct from the previous 4 new points.

In addition, each of the points which exist in the illustration of $T(4)$ (Diagram 2) continue to exist in the illustration of $T(5)$ (Diagram 3) and are distinct from each of the new points which were added.

Therefore, we get $$\begin{aligned} T(5)& =T(4)+4+2\\ & =14+4+2\\ & =20\\\end{aligned}$$

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Therefore, $T(4)=14$ and $T(5)=20$.
As in part (a), consider finding $T(n)$ with the help of (in terms of) $T(n-1)$ for any integer $n\geq2$.

To determine $T(n)$, add 1 line segment to any illustration of $T(n-1)$.

This new ($n^{th}$) line segment must intersect each of the existing $n-1$ line segments exactly once, creating $n-1$ new points of intersection.

This $n^{th}$ line segment also adds 2 new endpoints (distinct from the previous $n-1$ points).

In addition, each of the points which exist in the illustration of $T(n-1)$ continue to exist in the illustration of $T(n)$ and are distinct from each of the new points which were added.
Therefore, we get $T(n)=T(n-1)+(n-1)+2$ or $T(n)=T(n-1)+n+1$ and so $T(n)-T(n-1)=n+1$ for all $n\geq2$.
From part (b), $T(n)-T(n-1)=n+1$ and so $T(n)=T(n-1)+n+1$.

That is, the addition of an $n^{th}$ line segment increases $T(n-1)$ by $n+1$.

For example since $T(1)=2$, then $T(2)=T(1)+3=2+3$.

For small values of $n$, we determine $T(n)$ in the table below.

$n$
$T(n)=T(n-1)+n+1, n\geq2$

2
$T(2)=T(1)+3=2+3$

3
$T(3)=T(2)+4=2+3+4$

4
$T(4)=T(3)+5=2+3+4+5$

5
$T(5)=T(4)+6=2+3+4+5+6$

6
$T(6)=T(5)+7=2+3+4+5+6+7$

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We may use the pattern in the table above to establish an equation for $T(n)$.

What is the pattern?

Consider for example the row for $n=5$.

$T(5)$ is the sum of the positive integers from 2 to $n+1=5+1=6$.

This is true for each of the rows shown in the table.

That is, $T(n-1)=2+3+4+\dots+n$ for any positive integer $n\geq3$.

(Verify that this is true for each of the rows shown in the table.)

Since the addition of an $n^{th}$ line segment increases $T(n-1)$ by $n+1$, then $T(n)=T(n-1)+n+1$ and so $T(n)=(2+3+4+\dots+n)+n+1$.

Reorganizing this equation for $T(n)$, we get $$T(n)=1+2+3+4+\cdots+n+n.$$ Since the sum of the first $n$ positive integers $1+2+3+4+\cdots+n$ is equal to $\dfrac{n(n+1)}{2}$, then $T(n)=\dfrac{n(n+1)}{2}+n$.

Solving $T(n)=2015$, we get

$$\begin{aligned} \dfrac{n(n+1)}{2}+n& =2015\\ n(n+1)+2n& =4030\\ n^2+3n& =4030\\ n^2+3n-4030& =0\\ (n-62)(n+65)& =0\end{aligned}$$ and so $n=62$ (since $n>0$). (We could use the quadratic formula if we didn’t see how to factor the quadratic.)
Therefore, $n=62$ is the only value of $n$ for which $T(n)=2015$.