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Derive a formula for the series

$f(n)=1+1+4+1+4+9+.....+1+4+9+16...n^2$

Example

$f(3)= 1+1+4+1+4+9$

I couldn't really find any derivation of this online or a question like this it self.

$f(n)=displaystylesum_{k=1}^n dfrac{k(k+1)(2k+1)}{6}=sum_{k=1}^n dfrac{2k^3+3k^2+k}{6}=dfrac{1}{6}left(2sum_{k=1}^n k^3+3sum_{k=1}^nk^2+sum_{k=1}^nkright)$.

(Assuming you know the formula for the sum of cubes and squares of first $n$ natural numbers.)

As suggested by the user Yadati Kiran, $$f(n)=frac {1}{12} n(n+1)(n^2+3n+2)$$

Hint: This is just
$$sum_{k=1}^n left(sum_{j=1}^k j^2right)$$

With application of hockeystick identity:

$$begin{aligned}sum_{k=1}^{n}sum_{m=1}^{k}m^{2} & =sum_{k=1}^{n}sum_{m=1}^{k}left[2binom{m}{2}+binom{m}{1}right]\
& =2sum_{k=1}^{n}sum_{m=1}^{k}binom{m}{2}+sum_{k=1}^{n}sum_{m=1}^{k}binom{m}{1}\
& =2sum_{k=1}^{n}binom{k+1}{3}+sum_{k=1}^{n}binom{k+1}{2}\
& =2binom{n+2}{4}+binom{n+2}{3}\
& =frac{1}{12}left(n+2right)left(n+1right)nleft(n-1right)+frac{1}{6}left(n+2right)left(n+1right)n\
& =frac{1}{12}left(n+2right)left(n+1right)^{2}n
end{aligned}
$$