Pascal's hockey shot

Relevant wiki: Hockey Stick Identity

There is a pattern in Pascal's Triangle called the " Hockey Stick Pattern ," which is based on the Hockey Stick Identity : $\displaystyle\sum\limits_{k=r}^{n}\binom{k}{r}=\binom{n+1}{r+1}$ .

If you sum all the elements in red, then the sum will be the next element down diagonally in the opposite direction. If you are familiar with the game of hockey, you will recognize that the circled numbers resembles a hockey stick:

The answer is $\boxed{2600}$ , which is equal to $\binom{25+1}{2+1}$ .

My first solution on this website, also algebraic:

Notice that: 1, 2, 3, 6, 10...

Is equal to: 1+0, 1+2, 3+3, 6+4, 10+5... etc.

Here we can see that each number appears twice, so we can also write everything as:

2 x 1 + 2 which are the first 2 numbers 2 x 6 + 4, 2 x 15 + 6,
until 2 x 276 + 24

Notice that these numbers are the numbers we add but squared So they are: 2^2 4^2 6^2 8^2 etc.

So we can write them as the sum of (2n)^2 from 1 to 12 or 4n^2. And this is equal to: 4k(k+1)(2k+1)/6

We put 12 into the equation and get 2600

Without a combinatorial approach, we see that the sum of two consecutive numbers in the triangle is equal to the number directly below them, so in the figure, we see that the first two elements, $1+3$ sum up to $4$ , which is directly below $3$ . Similarly, $4+6$ sums up to $10$ , which is directly below $6$ . Continuing this, we finally see that $300+2300$ sums up to $\boxed{2600}$ , which is directly below $300$ , and, quite coincidentally, is our final answer.

P.S. Sorry if I have no visuals, as I am not yet that well versed in LaTeX.

An algebraic approach:

Notice that

$1, \; 1+2=3,\; 1+2+3=6 \ldots$

From here, we can deduce that each highlighted term, $T_n$ , represents the sum of the first $n$ terms of the arithmetic progression $1,2,3,\ldots$

We can also say that $T_n = \displaystyle \sum_{k=1}^n k = \dfrac{n(n+1)}{2}$

There are $24$ terms from the $2^{\text{nd}}$ row to the $25^{\text{th}}$ row, and the sum of all these terms are

$S_{24} \\ = \displaystyle \sum_{k=1}^{24} \dfrac{k(k+1)}{2}\\ =\dfrac{1}{2}\displaystyle\sum_{k=1}^{24}(k^2+k)\\ =\dfrac{1}{2}\left(\dfrac{24(25)(49)}{6} + \dfrac{24(25)}{2}\right)\\ =\dfrac{1}{2}\left(4900 +300\right)\\ =\dfrac{5200}{2}\\ =\boxed{2600}$

We know that each row in the Pascal's triangle has the coefficients for the expression of (x+1)^n where n runs from 0 to \infty. From the highlighted terms, we can thus infer that they are the successive terms of the sequence n C 2 added from 1 to 25. Doing some simple summation we arrive at \boxed{2600}.