An interesting little tidbit from Tanya Khovanova: consider the Pythagorean triple 32 + 42 = 52, in which all three numbers are consecutive, and another similar coincidental-seeming equation 102 + 112 + 122 = 132 + 142. These are the first two elements of a sequence of similar equations, with k+1 squares on the left-hand side and k squares on the right-hand side, which she presents there.
Still perhaps coincidental is the fact that 32 + 42 = 52 and 33 + 43 + 53 = 63. Of course once you see these it's tempting to check if 34 + 44 + 54 + 64 = 74. (It doesn't.) In fact, for n ≥ 4,
3n + 4n + ... + (n+2)n < (n+3)n
and here's a proof. I'll prove it for n ≥ 5. It suffices to show that
(3/(n+3))n + (4/(n+3))n + ... + ((n+2)/(n+3))n < 1.
Call the left-hand side f(n). But the last term on the right-hand side is decreasing in n, and for n ≥ 5, is at most 16807/32768 (its value at 5). Now, we can write the left-hand side of the above inequality as
((n+2)/(n+3))n (1 + ((n+1)/(n+2))n + (n/(n+2))n + ... + (3/(n+2))n)
and each term in the second factor is at most ((n+1)/(n+2))n times the previous one. So we have
(1 + ((n+1)/(n+2))n + (n/(n+2))n + ... + (3/(n+2))n) > (1 + r + r2 + r3 + ...)
where r = ((n+1)/(n+2))n. Of course the right-hand side of the previous inequality is 1/(1-r). r is positive and decreasing in n, so 1/(1-r) is as well. Since n ≥ 5, 1/(1-r) is bounded above by its value at 5, which is 16807/9031. Thus for n ≥ 5,
f(n) < (16807/32768)(16807/9031)
and the right-hand side is easily checked to be less than 1.
For n = 4 this proof doesn't work, because I threw away a bit too much in bounding the sum of a finite series by the sum of an infinite one, but it's easy to check.
In fact, f(n) approaches 1/(e-1) as n goes to infinity.
Edit (8:41 pm):: See also The Everything Seminar.