(i) (3+5-7+9+11-13+15+17-19+\cdots) to (3n) terms
Group in threes: ((3+5-7)+(9+11-13)+(15+17-19)+\cdots)
Each group: (1,\ 7,\ 13,\ldots) which is A.P. with first term 1, common difference 6.
Sum of (n) groups: [ S = \dfrac{n}{2}\bigl[2\cdot1+(n-1)6\bigr] = \dfrac{n}{2}(2+6n-6)=\dfrac{n}{2}(6n-4)=n(3n-2) ]
Answer: (n(3n-2))