(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))

(ii) (1+4-7+10+13-16+19+22-25+\cdots) to (3n) terms

Groups: ((1+4-7)+(10+13-16)+(19+22-25)+\cdots = (-2)+7+16+\cdots)

A.P. with first term (-2), common difference (9).

[ S=\dfrac{n}{2}\bigl[-4+(n-1)9\bigr]=\dfrac{n}{2}(9n-13) ]

Answer: (\dfrac{n}{2}(9n-13))