Sum the array times n, except the last
J, 3 bytes
That was fun to find.
&amp;+/

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How it works
10 (&amp;+/) 3 1 4 1 5 will bind 10 as an argument of + as 10&amp;+, one verb that gets inserted between the elements of the list by /. So we have: 3 (10&amp;+) 1 (10&amp;+) 4 (10&amp;+) 1 (10&amp;+) 5. Now x n&amp;v y means that y gets applied to n&amp;v for x times. With J's right to left evaluation we get: to 5 add 1 times 10, add 4 times 10, add 1 times 10, add 3 times 10. A challenge made for J's stranger parts. :-) And because + is commutative, +&amp;/ would also be a valid solution.