Let f(n) = 2n + 5. Then we must simply calculate (f1), f(2), ..., (f12). And we have that f(1) = 2*1 + 5 <=> f(1) = 7; f(2) = 2*2 + 5 <=> f(2) = 9; f(3) = 2*3 + 5 <=> f(3) = 11. And so on. It's easy to see that the rest of the results will be 13, 15, 17, 19, 21 and so on. So the summation will be S = 7 + 9 + 11 +... + 29 <=> S = 216