>>51062 Well, you write down the first numbers (note the different colors of the balls): 1 1+2 = 3 1+2+3 = 6 1+2+3+4 = 10 So the 50th triangular number is the sum of the first 50 natural numbers. If you don't know that this is equal to n choose 2 = n(n+1)/2 for n=50 you can prove it as follows just like Gauss:
Write the numbers that you want to sum in ascending order and below in descending order: 1 2 3 ... n-2 n-1 n n n-1 n-2 ... 3 2 1
The sum in a column is always n+1 so the sum of all 2n numbers is n(n+1). Since you only want to count each number once, you need to divide by 2:
>>51081 You can also do this with the balls if you have problems with the equation:
Bring the Tn balls in a right triangle pattern: * ** *** ****
Use another triangle with the same number of balls, but in reverse order, and put the two of them together:
* **** ** *** *** ** **** * You obtain a rectangle with n rows and n+1 balls in each row, that is, n(n+1) balls in total. Since you used two triangles, the number of balls in Tn is half this amount, n(n+1)/2.
>>51095 It's only the application of the triangular numbers before. You have numbered balls / squares, so in the first rows: 1 23 456 and so on. Note that the rightmost number is equal to T_n for n=1,2,3...
Since in the problem you have 60 completely filled rows you have used the balls numbered from 1 to T(60).
In the 61st row are the balls numbered T(60)+1,..., T(60)+23 (which you need), and so on.
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