Well if you want to approximate it to a finite number of terms you can do it in Matlab without toolboxes or loops:
sumCos = @(x, n)(sum(((-1).^(0:n)).*(x.^(2*(0:n)))./(factorial(2*(0:n)))));
and then use it like this
sumCos(pi, 30)
The first parameter is the angle, the second is the number of terms you want to take the series to (i.e. effects the precision). This is a numerical solution which I think is really what you're after.
btw I took the liberty of correcting your initial sum, surely n must start from 0 if you are trying to approximate cos
If you want to understand my formula (which surely you do) then you need to read up on some essential Matlab basics namely the colon operator and then the concept of using .
to perform element-wise operations.