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Originally Posted by Solo CarrieraWe will do this in the case of n = 5 for concreteness. From Eq. (1.4) we want
to sum all the numbers in a triangle:
1
1 + 2
1 + 2 + 3
1 + 2 + 3 + 4
1 + 2 + 3 + 4 + 5
The kth row is the triangular number tk . We take three copies of the triangle,
each one rotated by 120◦:
1 1 5
1+2 2+1 4+ 4
1 + 2+3 3+2 +1 3+ 3+ 3
1 + 2+ 3+4 4+3 +2 +1 2+ 2+ 2+ 2
1 + 2+ 3+ 4+5 5+ 4 +3 +2 +1 1+ 1+ 1+ 1+1
The rearranged triangles still have the same sum. This is the analog of Gauss
taking a second copy of the sum for tn written backward. Observe that if we
add the left and center triangles together, in each row the sums are constant:
1 + 1 = 2
1 + 2 + 2 + 1 = 3 + 3
1 + 2 +3 + 3 + 2 + 1 = 4 + 4 + 4
1 + 2 + 3 + 4 + 4 + 3 + 2 + 1 = 5 + 5 + 5 + 5
1 + 2 + 3 + 4 + 5 + 5 + 4 + 3 + 2 + 1 = 6 + 6 + 6 + 6 + 6
I n row k, all the entries are k + 1, just as Gauss found. In the third triangle,
all the entries in row k are the same; they are equal to n − k + 1, and k + 1
plus n − k + 1 is n + 2.
2 + 5 = 7
3 + 3 + 4 + 4 = 7 + 7
4 + 4 + 4 + 3 + 3 + 3 = 7 + 7 + 7
5 + 5 + 5 + 5 + 2 + 2 + 2 + 2 = 7 + 7 + 7 + 7
6 + 6 + 6 + 6 + 6 + 1 + 1 + 1 + 1 + 1 = 7 + 7 + 7 + 7 + 7
We get a triangle with tn numbers in it, each of which is equal to n + 2. So,
3Tn = tn(n + 2) = n(n + 1)(n + 2)/2,
10 1. Sums and Differences
Table 1.1. Another proof of
Nicomachus identity
1 2 3 4 5 . . .
2 4 6 8 10 . . .
3 6 9 12 15 . . .
4 8 12 16 20 . . .
5 10 15 20 25 . . .
...
...
...
...
...
is true and use it to prove the next case. But
13 + 23 +· · ·+(n − 1)3 + n3
={13 + 23 + ·· ·+(n − 1)3} + n3
=
(n − 1)2n2
4 + n3
by the induction hypothesis. Now, put the two terms over the common denominator
and simplify to get n2(n + 1)2/4.
Exercise 1.1.5. Here’s another proof that
13 + 23 + 33 + ·· ·+n3 = n2(n + 1)2/4, (1.9)
