How do I find the sum of arithmetic series? [duplicate]

I think that, unless we are told anything about the relationship between the costs of a drumming drummer, a piping piper, a leaping lord, a dancing lady, a milking maid, a swimming swan, a laying goose, a golden ring, a calling bird, a french hen, a turtle dove and a partridge, then there's not much simplification to be done.

Arthur is right - the relationship between $c_n$ is required in order to simplify further.

If, instead, $c_n=1$ (all objects cost $1), the total price is

$$\sum_{n=1}^{12}n(13-n)=\sum_{n=1}^{12}\sum_{m=n}^{12}n=\sum_{m=1}^{12}\sum_{n=1}^m \binom n1=\sum_{m=1}^{12}\binom {m+1}2=\binom{14}3=364\quad\blacksquare$$

This also means that the total number of objects is $364$.

Curiously and coincidentally, this also means that there is one gift for each day of the year*, apart from one day - perhaps Christmas Eve!

*assuming a non-leap year