A number of the form 1 + 2 + 3 + .. - Leaving Cert Mathematics - Question 7 - 2020

To determine if 1275 is a triangular number, we can set up the equation:

$\frac{n(n+1)}{2} = 1275$

Multiplying both sides by 2 gives:

$n(n+1) = 2550$

Rearranging, we get:

$n^2 + n - 2550 = 0$

Using the quadratic formula:

$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

i.e. $n = \frac{-1 \pm \sqrt{1 + 10200}}{2} = \frac{-1 \pm 101}{2}$

This yields two solutions, $n = 50$ and a negative solution, thus 1275 is the 50th triangular number.

$T_{n + 1} = \frac{n(n+1)}{2} + (n + 1) = \frac{n(n+1) + 2(n + 1)}{2} = \frac{n^2 + n + 2n + 2}{2} = \frac{n^2 + 3n + 2}{2} = \frac{(n+1)(n + 2)}{2}$

Let the two consecutive triangular numbers be $T_n$ and $T_{n+1}$.

Then:

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

Simplifying,

$= \frac{n(n+1) + (n+1)(n+2)}{2} = \frac{(n + 1)(n + n + 2)}{2} = \frac{(n + 1)(2n + 2)}{2} = (n + 1)(n + 1) = (n + 1)^2$

Thus, the sum is always a square number.

Let the consecutive triangular numbers be $T_n$ and $T_{n + 1}$.

Thus, we have:

$T_n + T_{n + 1} = 12544$

Substituting for triangular numbers:

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

And conducting a similar calculation as earlier, we arrive at: $n$ should be around 111, so:

$T_{111} = \frac{111(112)}{2} = 6216$

Thus, the smaller number is 6216.

Using the formula:

$N_k = \frac{(3 + 2\sqrt{2})^k - (3 - 2\sqrt{2})^k}{4\sqrt{2}}$

Calculating for $k = 3$:

$N_3 = \frac{(3 + 2\sqrt{2})^3 - (3 - 2\sqrt{2})^3}{4\sqrt{2}} = 1225$