(i) ( 1\times 3 + 2\times 5 + 3\times 7 + \cdots )

The ( n^{th} ) term is [ u_n = n(2n+1) = 2n^2 + n. ]

Sum to ( n ) terms: [ S_n = \sum_{k=1}^{n} (2k^2 + k) = 2\cdot\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} = \frac{n(n+1)}{6}\bigl[2(2n+1) + 3\bigr] = \frac{n(n+1)(4n+5)}{6}. ]

Answer: ( \dfrac{n(n+1)(4n+5)}{6} )

(ii) ( 1\times 5 + 2\times 8 + 3\times 11 + \cdots )

[ u_n = n(3n+2) = 3n^2 + 2n. ]

[ S_n = 3\cdot\frac{n(n+1)(2n+1)}{6} + 2\cdot\frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{2} + n(n+1) = \frac{n(n+1)}{2}\bigl[(2n+1) + 2\bigr] = \frac{n(n+1)(2n+3)}{2}. ]

Answer: ( \dfrac{n(n+1)(2n+3)}{2} )

(iii) ( 1\times 2 + 2\times 5 + 3\times 8 + \cdots )

[ u_n = n(3n-1) = 3n^2 - n. ]

[ S_n = 3\cdot\frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{2} - \frac{n(n+1)}{2} = \frac{n(n+1)}{2}\bigl(2n+1-1\bigr) = n^2(n+1). ]

Answer: ( n^2(n+1) )

(iv) ( 1\times 3\times 5 + 2\times 4\times 6 + 3\times 5\times 7 + \cdots )

[ u_n = n(n+2)(n+4) = n(n^2 + 6n + 8) = n^3 + 6n^2 + 8n. ]

[ S_n = \sum k^3 + 6\sum k^2 + 8\sum k = \left(\frac{n(n+1)}{2}\right)^2 + 6\cdot\frac{n(n+1)(2n+1)}{6} + 8\cdot\frac{n(n+1)}{2} = \frac{n^2(n+1)^2}{4} + n(n+1)(2n+1) + 4n(n+1) = \frac{n(n+1)}{4}\Bigl[n(n+1) + 4(2n+1) + 16\Bigr] = \frac{n(n+1)}{4}(n^2 + n + 8n + 4 + 16) = \frac{n(n+1)}{4}(n^2 + 9n + 20) = \frac{n(n+1)(n+4)(n+5)}{4}. ]

Answer: ( \dfrac{n(n+1)(n+4)(n+5)}{4} )

(v) ( 1\times 2\times 4 + 2\times 3\times 7 + 3\times 4\times 10 + \cdots )

[ u_n = n(n+1)(3n+1) = n(3n^2 + n + 3n + 1) = 3n^3 + 4n^2 + n. ]

[ S_n = 3\sum k^3 + 4\sum k^2 + \sum k = 3\left(\frac{n(n+1)}{2}\right)^2 + 4\cdot\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{3n^2(n+1)^2}{4} + \frac{2n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} = \frac{n(n+1)}{12}\Bigl[9n(n+1) + 8(2n+1) + 6\Bigr] = \frac{n(n+1)}{12}(9n^2 + 9n + 16n + 8 + 6) = \frac{n(n+1)}{12}(9n^2 + 25n + 14) = \frac{n(n+1)(9n^2 + 25n + 14)}{12}. ]

Answer: ( \dfrac{n(n+1)(9n^2 + 25n + 14)}{12} )

(vi) ( 2^2 + 4^2 + 6^2 + \cdots )

[ u_n = (2n)^2 = 4n^2, \qquad S_n = 4\cdot\frac{n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3}. ]

Answer: ( \dfrac{2n(n+1)(2n+1)}{3} )

(vii) ( 3^2 + 6^2 + 9^2 + \cdots )

[ u_n = (3n)^2 = 9n^2, \qquad S_n = 9\cdot\frac{n(n+1)(2n+1)}{6} = \frac{3n(n+1)(2n+1)}{2}. ]

Answer: ( \dfrac{3n(n+1)(2n+1)}{2} )

(viii) ( 4\times 1^2 + 7\times 2^2 + 10\times 3^2 + \cdots )

[ u_n = (3n+1)n^2 = 3n^3 + n^2. ]

[ S_n = 3\left(\frac{n(n+1)}{2}\right)^2 + \frac{n(n+1)(2n+1)}{6} = \frac{3n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)}{12}\Bigl[9n(n+1) + 2(2n+1)\Bigr] = \frac{n(n+1)}{12}(9n^2 + 9n + 4n + 2) = \frac{n(n+1)(9n^2 + 13n + 2)}{12}. ]

Answer: ( \dfrac{n(n+1)(9n^2 + 13n + 2)}{12} )

(ix) ( 3 + (3+7) + (3+7+11) + \cdots )

The ( r^{th} ) group is the sum of an A.P. with first term 3 and common difference 4, having ( r ) terms: [ s_r = \frac{r}{2}\bigl[2\cdot 3 + (r-1)4\bigr] = \frac{r}{2}(6 + 4r - 4) = \frac{r}{2}(4r + 2) = r(2r + 1) = 2r^2 + r. ]

Thus [ S_n = \sum_{r=1}^{n} (2r^2 + r) = 2\cdot\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} = \frac{n(n+1)}{6}(4n + 2 + 3) = \frac{n(n+1)(4n+5)}{6}. ]

Answer: ( \dfrac{n(n+1)(4n+5)}{6} )

(x) ( 1^2 + (1^2+2^2) + (1^2+2^2+3^2) + \cdots )

The ( r^{th} ) group is [ s_r = \sum_{k=1}^{r} k^2 = \frac{r(r+1)(2r+1)}{6}. ]

[ S_n = \sum_{r=1}^{n} \frac{r(r+1)(2r+1)}{6} = \frac{1}{6}\sum (2r^3 + 3r^2 + r) = \frac{1}{6}\Bigl[2\left(\frac{n(n+1)}{2}\right)^2 + 3\cdot\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\Bigr] = \frac{1}{6}\Bigl[\frac{n^2(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{2} + \frac{n(n+1)}{2}\Bigr] = \frac{n(n+1)}{12}\Bigl[n(n+1) + (2n+1) + 1\Bigr] = \frac{n(n+1)}{12}(n^2 + n + 2n + 1 + 1) = \frac{n(n+1)(n^2 + 3n + 2)}{12} = \frac{n(n+1)^2(n+2)}{12}. ]

Answer: ( \dfrac{n(n+1)^2(n+2)}{12} )