Square pyramidal numbers

Python 2, 22 bytes

lambda n:n*~n*~(n*2)/6

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This uses the closed-form formula n * (n+1) * (2*n+1) / 6. The code performs the following operations:

• Multiplies n by (n*):

  • The bitwise complement of n (~n), which essentially means -1-n.
  • And by the bitwise complement of 2n (*~(n*2)), which means -1-2n.

• Divides by 6 (/6).

Python 2, 27 bytes

f=lambda n:n and f(n-1)+n*n

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_{Saved 1 byte thanks to Rod and 1 thanks to G B.}

MATL, 3 bytes

:Us

... or them?

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Explanation

:    % Implicit input n. Push range [1 2 ... n]
U    % Square, element-wise
s    % Sum of array. Implicit display

JavaScript (ES6), 16 bytes

n=>n*(n+++n)*n/6

Demo

let f =

n=>n*(n+++n)*n/6

for(n = 0; n < 10; n++) {
  console.log('a(' + n + ') = ' + f(n))
}

How?

The expression n+++n is parsed as n++ + n ⁽¹⁾. Not that it really matters because n + ++n would also work in this case.

Therefore:

n*(n+++n)*n/6 =
n * (n + (n + 1)) * (n + 1) / 6 =
n * (2 * n + 1) * (n + 1) / 6

which evaluates to sum(k=0...n)(k²).

_{(1) This can be verified by doing n='2';console.log(n+++n) which gives the integer 5, whereas n + ++n would give the string '23'.}