For any integer n 0, the triangular number Tn is the number of dots in a triangular array with n points on each side. What is the value of the 64th triangle number?

Question

For any integer n > 0, the triangular number Tn is the number of dots in a triangular array with n points on each side. What is the value of the 64th triangle number?

stamp · Answer

$$T_1=1, \ T_2=3,\ T_3=6, \ T_4=10$$$$Find \ T_{64}.$$

anonymous · Answer

Note that T(1)=1, T(2)+1+2, T(3)=1+2+3...
So we have T(n)=1+2+3+4+...+(n-1)+n

stamp · Answer

@Xavier Correct. Where do we go from here?

anonymous · Answer

This is generally a well known sum that evaluates to n(n+1)/2
The way this can be shown is that you can make n/2 (n+1)s by pairing the last and the first number, then the second and the second last number etc.
So:
n+1
2+(n-1)
3+(n-3)
...
We can do this n/2 times so the sum evaluates to n(n+1)/2

Or it can be proven inductively I guess

anonymous · Answer

Sorry the second one in that small list should be 1+(n-1) then 2+(n-2)

stamp · Answer

@Xavier You are correct. I was considering this problem earlier and could not come up with a solution. I referenced some resources and ended up finding an approach, but it was not as ideal as yours. Thank you for your input.

anonymous · Answer

Glad to help.

whpalmer4 · Answer

Here's another way of looking at this: Your triangle with $n$ dots on a side is also $n$ layers tall. If you take another such triangle (same value of $n$), flip it upside down, and nestle the two together side-by-side, you have a rectangular array of dots which is $n*(n+1)$ dots in total. The triangular number is half of those dots, so $$T_n = \frac{n(n+1)}{2}$$ @Xavier's explanation is the same trick that Gauss used to quickly sum the numbers from 1 to 100 as a child. Read a hundred or so versions of the story at http://www.sigmaxi.org/amscionline/gauss-snippets.html