Although you probably have not learned it yet, there is a general method for problems like this: Newton's divided difference method.
Set up a diagram like this:
[tex]\begin{array}{ccccc}n & f(n) & \Delta f & \Delta^2 f & \Delta^3 f \\0 & 0 & 2 & 3 & 1 \\ 1 & 2& 5 & 4 & 1 \\ 2 & 7 & 9 & 5 & 1 \\ 3 & 16 & 14 & 6 & _ \\ 4 & 30 & 20 & _ & _ \\ 5 & 50 & _ & _ & _\end{array}[/tex]
That is, the first column is the non-negative integers, the second column, f(n), is the partial sum up to that n, the third column is the difference between two consecutive numbers in the second column (and is just the individual terms in the series), the fourth column is the difference between two consecutive numbers in the third column, and the fifth column is the difference between two consecutive numbers in the fourth column. We can stop there because all numbers in the fourth column are the same so all succeeding columns will be all zeros. I put in the n= 0 row by working backward from the last column.
Now, Newton's divided difference formula is an analog of Taylor's series for continuous functions, using "[itex]\Delta[/itex]" instead of the derivative:
[tex]f(n)= f(0)+ \Delta f(0) n+ \frac{\Delta^2 f(0)}{2!}n(n-1)+ \frac{\Delta^3 f(0)}{3!}n(n-1)(n-2)[/itex]<br />
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For this particular sequence we have <br />
f(n)= 0+ 2n+ (3/2)n(n-1)+ (1/6)n(n-1)(n-2)<br />
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Multiply that out and you will get a cubic formula (or product of three linear terms) that will give the nth sum.[/tex]
