Why does +/@*: 1 2 3 give me 1 4 9 instead of 14?
+/ *: 1 2 3 14
and the Dictionary says that u@v y is u v y.
So how come?
+/@*: 1 2 3 1 4 9
The problem is the rank of the verb. u@v creates a verb whose rank is the rank of v. This verb is equivalent to u v y but it is applied to each cell of y independently. The rank of +/@*: is the same as the rank of *:, i. e. 0; so +/ *: y happens for each scalar in y independently, rather than summing across the scalars.
The solution is to make sure the combined verb has infinite rank, so it will be applied to the entire y operand. The preferred way to do that is by using @: rather than @ as in
+/@:*: 1 2 3 14
Contributed by HenryRich