Find the sum to n terms of each of the series.
1. $\displaystyle 1\times 2+2\times 3+3\times 4+4\times 5+...$
Given the series $\displaystyle 1\times 2+2\times 3+3\times 4+4\times 5+...$
⇒ $\displaystyle n^{th}$ term is $\displaystyle a_{n}=n(n+1)$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}$ $\displaystyle =\sum_{k=1}^{n}k(k+1)$
$\displaystyle =\sum_{k=1}^{n}(k^{2}+k)$
$\displaystyle =\sum_{k=1}^{n}k^{2}+\sum_{k=1}^{n}k$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$
$\displaystyle =\frac{n(n+1)(2n+1)+3n(n+1)}{6}$
$\displaystyle =\frac{n(n+1)(2n+1+3)}{6}$
$\displaystyle =\frac{n(n+1)(2n+4)}{6}$
$\displaystyle =\frac{2n(n+1)(n+2)}{6}$
$\displaystyle =\frac{n(n+1)(n+2)}{3}$
2. $\displaystyle 1\times 2\times 3+2\times 3\times 4$ $\displaystyle +3\times 4\times 5+...$
Given the series $\displaystyle 1\times 2\times 3+2\times 3\times 4+3\times 4\times 5+...$
$\displaystyle n^{th}$ term is $\displaystyle a_{n}=n(n+1)(n+2)$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}$ $\displaystyle =\sum_{k=1}^{n}k(k+1)(k+2)$
$\displaystyle =\sum_{k=1}^{n}(k^{3}+3k^{2}+2k)$
$\displaystyle =\sum_{k=1}^{n}k^{3}+2\sum_{k=1}^{n}k^{2}+2\sum_{k=1}^{n}k$
$\displaystyle =\frac{[n(n+1)]^{2}}{4}+3\frac{n(n+1)(2n+1)}{6}$ $\displaystyle +2\frac{n(n+1)}{2}$
$\displaystyle =\frac{[n(n+1)]^{2}}{4}+\frac{n(n+1)(2n+1)}{2}$ $\displaystyle +n(n+1)$
$\displaystyle =n(n+1)$ $\displaystyle \left [\frac{n(n+1)}{4}+\frac{(2n+1)}{2}+1 \right ]$
$\displaystyle =n(n+1)$ $\displaystyle \left [\frac{n(n+1)+2(2n+1)+4}{4} \right ]$
$\displaystyle =n(n+1)\left [\frac{n^{2}+5n+6}{4} \right ]$
$\displaystyle =\frac{n(n+1)(n+2)(n+3)}{4}$
3. $\displaystyle 3\times 1^{2}+5\times 2^{2}+7\times 3^{2}+...$
Given the series $\displaystyle 3\times 1^{2}+5\times 2^{2}+7\times 3^{2}+...$
$\displaystyle n^{th}$ term is $\displaystyle a_{n}=(2n+1)n^{2}$
$\displaystyle S_{n}=\sum_{k=1}^{n}(2k+1)k^{2}$
$\displaystyle =\sum_{k=1}^{n}(2k^{3}+k^{2})$
$\displaystyle =2\sum_{k=1}^{n}k^{3}+\sum_{k=1}^{n}k^{2}$
$\displaystyle =2\frac{[n(n+1)]^{2}}{4}+\frac{n(n+1)(2n+1)}{6}$
$\displaystyle =\frac{[n(n+1)]^{2}}{2}+\frac{n(n+1)(2n+1)}{6}$
$\displaystyle =n(n+1)\left [\frac{n(n+1)}{2}+\frac{(2n+1)}{6} \right ]$
$\displaystyle =n(n+1)\left [\frac{3n(n+1)+2n+1}{6} \right ]$
$\displaystyle =\frac{n(n+1)(3n^{2}+5n+1)}{6}$
4. $\displaystyle \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...$
Given the series, $\displaystyle \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...$
$\displaystyle a_{n}=\frac{1}{n(n+1)}$ ⇒
$\displaystyle a_{n}=\frac{1}{n}-\frac{1}{n+1}$
$\displaystyle a_{1}=\frac{1}{1}-\frac{1}{2}$
$\displaystyle a_{2}=\frac{1}{2}-\frac{1}{3}$
$\displaystyle a_{3}=\frac{1}{3}-\frac{1}{4}$
$\displaystyle a_{n}=\frac{1}{n}-\frac{1}{n+1}$
$\displaystyle S_{n}=a_{1}+a_{2}+a_{3}+...+a_{n}$
$\displaystyle =\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}$
$\displaystyle =\frac{1}{1}-\frac{1}{n+1}$
$\displaystyle =\frac{n+1-1}{n+1}=\frac{n}{n+1}$
5. $\displaystyle 5^{2}+6^{2}+7^{2}+...+20^{2}$
$\displaystyle [1^{2}+2^{2}+3^{2}+4^{2}+...+20^{2}]$ - $\displaystyle [1^{2}+2^{2}+3^{2}+4^{2}]$
$\displaystyle S_{n}=\sum_{k=1}^{20}k^{2}-\sum_{k=1}^{4}k^{2}$ [$\displaystyle \sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}$]
$\displaystyle S_{n}=\frac{20(20+1)(2(20)+1)}{6}$ $\displaystyle -\frac{4(4+1)(2(4)+1)}{6}$
$\displaystyle =\frac{20(21)(41)}{6}-\frac{4(5)(9)}{6}$
$\displaystyle =\frac{17220}{6}-\frac{180}{6}=2840$
6. $\displaystyle 3\times 8+6\times 11+9\times 14+...$
$\displaystyle =(n^{th}\; term\; of \; 3,6,9,...)\times $ $\displaystyle (n^{th}\; term\; of \; 8,11,14,...)$
$\displaystyle =(3n)\times (3n+5)$
$\displaystyle =9n^{2}+15n$
$\displaystyle S_{n}=\sum_{k=1}^{n}[9k^{2}+15k]$
$\displaystyle S_{n}=9\sum_{k=1}^{n}k^{2}+15\sum_{k=1}^{n}k$
$\displaystyle =9\left [ \frac{n(n+1)(2n+1)}{6} \right ]$ $\displaystyle +15\left [ \frac{n(n+1)}{2} \right ]$
$\displaystyle =\left [ \frac{3n(n+1)(2n+1)}{2} \right ]$ $\displaystyle +\left [ \frac{15n(n+1)}{2} \right ]$
$\displaystyle =\frac{3n(n+1)}{2}(2n+1+5)$ $\displaystyle =\frac{3n(n+1)}{2}(2n+6)$
$\displaystyle =3n(n+1)(n+3)$
7. $\displaystyle 1^{2}+(1^{2}+2^{2})+(1^{2}+2^{2}+3^{2})$ $\displaystyle +...$
$\displaystyle =1^{2}+2^{2}+3^{2}+...+n^{2}$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}$
$\displaystyle =\frac{(n^{2}+n)(2n+1)}{6}$
$\displaystyle =\frac{(2n^{3}+n^{2}+2n^{2}+n)}{6}$
$\displaystyle =\frac{(2n^{3}+3n^{2}+n)}{6}$
$\displaystyle =\frac{1}{3}n^{3}+\frac{1}{2}n^{2}+\frac{1}{6}n$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}$ $\displaystyle =\sum_{k=1}^{n}\left [ \frac{1}{3} k^{3}+\frac{1}{2} k^{2}+\frac{1}{6} k\right ]$
$\displaystyle =\frac{1}{3}\sum_{k=1}^{n}k^{3}+\frac{1}{2}\sum_{k=1}^{n}k^{2}+\frac{1}{6}\sum_{k=1}^{n}k$
$\displaystyle =\frac{1}{3}\left [ \frac{(n(n+1))^{2}}{4} \right ]$ $\displaystyle +\frac{1}{2}\left [ \frac{n(n+1)(2n+1)}{6} \right ]$ $\displaystyle +\frac{1}{6}\left [ \frac{n(n+1)}{2} \right ]$
$\displaystyle =\frac{n(n+1)}{12}$ $\displaystyle \left [ n(n+1)+(2n+1)+1 \right ]$
$\displaystyle =\frac{n(n+1)}{12}\left [ n^{2}+n+2n+1+1 \right ]$
$\displaystyle =\frac{n(n+1)}{12}\left [ n^{2}+3n+2 \right ]$
$\displaystyle =\frac{n(n+1)}{12}\left [ (n+1)(n+2) \right ]$
$\displaystyle =\frac{n(n+1)^{2}(n+2)}{12}$
Find the sum to n terms of the following series whose $\displaystyle n^{th}$ term is given by
8. $\displaystyle n(n+1)(n+4)$
$\displaystyle a_{n}=n(n+1)(n+4)$
$\displaystyle =n^{3}+5n^{2}+4n$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}$ $\displaystyle =\sum_{k=1}^{n}(k^{3}+5k^{2}+4k)$
$\displaystyle =\sum_{k=1}^{n}k^{3}+5\sum_{k=1}^{n}k^{2}+4\sum_{k=1}^{n}k$
$\displaystyle =\frac{(n(n+1))^{2}}{4}$ $\displaystyle +5\left [ \frac{n(n+1)(2n+1)}{6} \right ]$ $\displaystyle +4\left [ \frac{n(n+1)}{2} \right ]$
$\displaystyle =\frac{n(n+1)}{2}$ $\displaystyle \left [ \frac{n(n+1)}{2}+\frac{5(2n+1)}{3}+4 \right ]$
$\displaystyle =\frac{n(n+1)}{2}$ $\displaystyle \left [ \frac{3n(n+1)+10(2n+1)+24}{6} \right ]$
$\displaystyle =\frac{n(n+1)}{2}$ $\displaystyle \left [ \frac{3n^{2}+3n+20n+10+24}{6} \right ]$
$\displaystyle =\frac{n(n+1)}{2}\left [ \frac{3n^{2}+23n+34}{6} \right ]$
$\displaystyle =\frac{n(n+1)(3n^{2}+23n+34)}{12}$
9. $\displaystyle n^{2}+2^{n}$
$\displaystyle a_{n}=n^{2}+2^{n}$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}(k^{2}+2^{k})$
$\displaystyle =\sum_{k=1}^{n}k^{2}+\sum_{k=1}^{n}2^{k}$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}$ $\displaystyle +(2^{1}+2^{2}+2^{3}+...+2^{n})$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}+\frac{2(2^{n}-1)}{2-1}$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}+2(2^{n}-1)$
10. $\displaystyle (2n-1)^{2}$
$\displaystyle a_{n}=(2n-1)^{2}$
$\displaystyle a_{n}=4n^{2}-4n+1$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}$ $\displaystyle=\sum_{k=1}^{n}(4k^{2}-4k+1)$
$\displaystyle=4\sum_{k=1}^{n}k^{2}-4\sum_{k=1}^{n}k+\sum_{k=1}^{n}1$
$\displaystyle =4\left [ \frac{n(n+1)(2n+1)}{6} \right ]-4\left [ \frac{n(n+1)}{2} \right ]+n$
$\displaystyle =n\left [ \frac{2(n+1)(2n+1)}{3} -2(n+1)+1\right ]$
$\displaystyle =n\left [ \frac{4n^{2}+6n+2}{3} -2(n+1)+1\right ]$
$\displaystyle =n\left [ \frac{4n^{2}+6n+2-6n-6+3}{3}\right ]$
$\displaystyle =n\left [ \frac{4n^{2}-1}{3}\right ]$
$\displaystyle = \frac{n(2n+1)(2n-1)}{3}$
1. $\displaystyle 1\times 2+2\times 3+3\times 4+4\times 5+...$
Given the series $\displaystyle 1\times 2+2\times 3+3\times 4+4\times 5+...$
⇒ $\displaystyle n^{th}$ term is $\displaystyle a_{n}=n(n+1)$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}$ $\displaystyle =\sum_{k=1}^{n}k(k+1)$
$\displaystyle =\sum_{k=1}^{n}(k^{2}+k)$
$\displaystyle =\sum_{k=1}^{n}k^{2}+\sum_{k=1}^{n}k$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$
$\displaystyle =\frac{n(n+1)(2n+1)+3n(n+1)}{6}$
$\displaystyle =\frac{n(n+1)(2n+1+3)}{6}$
$\displaystyle =\frac{n(n+1)(2n+4)}{6}$
$\displaystyle =\frac{2n(n+1)(n+2)}{6}$
$\displaystyle =\frac{n(n+1)(n+2)}{3}$
2. $\displaystyle 1\times 2\times 3+2\times 3\times 4$ $\displaystyle +3\times 4\times 5+...$
Given the series $\displaystyle 1\times 2\times 3+2\times 3\times 4+3\times 4\times 5+...$
$\displaystyle n^{th}$ term is $\displaystyle a_{n}=n(n+1)(n+2)$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}$ $\displaystyle =\sum_{k=1}^{n}k(k+1)(k+2)$
$\displaystyle =\sum_{k=1}^{n}(k^{3}+3k^{2}+2k)$
$\displaystyle =\sum_{k=1}^{n}k^{3}+2\sum_{k=1}^{n}k^{2}+2\sum_{k=1}^{n}k$
$\displaystyle =\frac{[n(n+1)]^{2}}{4}+3\frac{n(n+1)(2n+1)}{6}$ $\displaystyle +2\frac{n(n+1)}{2}$
$\displaystyle =\frac{[n(n+1)]^{2}}{4}+\frac{n(n+1)(2n+1)}{2}$ $\displaystyle +n(n+1)$
$\displaystyle =n(n+1)$ $\displaystyle \left [\frac{n(n+1)}{4}+\frac{(2n+1)}{2}+1 \right ]$
$\displaystyle =n(n+1)$ $\displaystyle \left [\frac{n(n+1)+2(2n+1)+4}{4} \right ]$
$\displaystyle =n(n+1)\left [\frac{n^{2}+5n+6}{4} \right ]$
$\displaystyle =\frac{n(n+1)(n+2)(n+3)}{4}$
3. $\displaystyle 3\times 1^{2}+5\times 2^{2}+7\times 3^{2}+...$
Given the series $\displaystyle 3\times 1^{2}+5\times 2^{2}+7\times 3^{2}+...$
$\displaystyle n^{th}$ term is $\displaystyle a_{n}=(2n+1)n^{2}$
$\displaystyle S_{n}=\sum_{k=1}^{n}(2k+1)k^{2}$
$\displaystyle =\sum_{k=1}^{n}(2k^{3}+k^{2})$
$\displaystyle =2\sum_{k=1}^{n}k^{3}+\sum_{k=1}^{n}k^{2}$
$\displaystyle =2\frac{[n(n+1)]^{2}}{4}+\frac{n(n+1)(2n+1)}{6}$
$\displaystyle =\frac{[n(n+1)]^{2}}{2}+\frac{n(n+1)(2n+1)}{6}$
$\displaystyle =n(n+1)\left [\frac{n(n+1)}{2}+\frac{(2n+1)}{6} \right ]$
$\displaystyle =n(n+1)\left [\frac{3n(n+1)+2n+1}{6} \right ]$
$\displaystyle =\frac{n(n+1)(3n^{2}+5n+1)}{6}$
4. $\displaystyle \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...$
Given the series, $\displaystyle \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...$
$\displaystyle a_{n}=\frac{1}{n(n+1)}$ ⇒
$\displaystyle a_{n}=\frac{1}{n}-\frac{1}{n+1}$
$\displaystyle a_{1}=\frac{1}{1}-\frac{1}{2}$
$\displaystyle a_{2}=\frac{1}{2}-\frac{1}{3}$
$\displaystyle a_{3}=\frac{1}{3}-\frac{1}{4}$
$\displaystyle a_{n}=\frac{1}{n}-\frac{1}{n+1}$
$\displaystyle S_{n}=a_{1}+a_{2}+a_{3}+...+a_{n}$
$\displaystyle =\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}$
$\displaystyle =\frac{1}{1}-\frac{1}{n+1}$
$\displaystyle =\frac{n+1-1}{n+1}=\frac{n}{n+1}$
5. $\displaystyle 5^{2}+6^{2}+7^{2}+...+20^{2}$
$\displaystyle [1^{2}+2^{2}+3^{2}+4^{2}+...+20^{2}]$ - $\displaystyle [1^{2}+2^{2}+3^{2}+4^{2}]$
$\displaystyle S_{n}=\sum_{k=1}^{20}k^{2}-\sum_{k=1}^{4}k^{2}$ [$\displaystyle \sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}$]
$\displaystyle S_{n}=\frac{20(20+1)(2(20)+1)}{6}$ $\displaystyle -\frac{4(4+1)(2(4)+1)}{6}$
$\displaystyle =\frac{20(21)(41)}{6}-\frac{4(5)(9)}{6}$
$\displaystyle =\frac{17220}{6}-\frac{180}{6}=2840$
6. $\displaystyle 3\times 8+6\times 11+9\times 14+...$
$\displaystyle =(n^{th}\; term\; of \; 3,6,9,...)\times $ $\displaystyle (n^{th}\; term\; of \; 8,11,14,...)$
$\displaystyle =(3n)\times (3n+5)$
$\displaystyle =9n^{2}+15n$
$\displaystyle S_{n}=\sum_{k=1}^{n}[9k^{2}+15k]$
$\displaystyle S_{n}=9\sum_{k=1}^{n}k^{2}+15\sum_{k=1}^{n}k$
$\displaystyle =9\left [ \frac{n(n+1)(2n+1)}{6} \right ]$ $\displaystyle +15\left [ \frac{n(n+1)}{2} \right ]$
$\displaystyle =\left [ \frac{3n(n+1)(2n+1)}{2} \right ]$ $\displaystyle +\left [ \frac{15n(n+1)}{2} \right ]$
$\displaystyle =\frac{3n(n+1)}{2}(2n+1+5)$ $\displaystyle =\frac{3n(n+1)}{2}(2n+6)$
$\displaystyle =3n(n+1)(n+3)$
7. $\displaystyle 1^{2}+(1^{2}+2^{2})+(1^{2}+2^{2}+3^{2})$ $\displaystyle +...$
$\displaystyle =1^{2}+2^{2}+3^{2}+...+n^{2}$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}$
$\displaystyle =\frac{(n^{2}+n)(2n+1)}{6}$
$\displaystyle =\frac{(2n^{3}+n^{2}+2n^{2}+n)}{6}$
$\displaystyle =\frac{(2n^{3}+3n^{2}+n)}{6}$
$\displaystyle =\frac{1}{3}n^{3}+\frac{1}{2}n^{2}+\frac{1}{6}n$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}$ $\displaystyle =\sum_{k=1}^{n}\left [ \frac{1}{3} k^{3}+\frac{1}{2} k^{2}+\frac{1}{6} k\right ]$
$\displaystyle =\frac{1}{3}\sum_{k=1}^{n}k^{3}+\frac{1}{2}\sum_{k=1}^{n}k^{2}+\frac{1}{6}\sum_{k=1}^{n}k$
$\displaystyle =\frac{1}{3}\left [ \frac{(n(n+1))^{2}}{4} \right ]$ $\displaystyle +\frac{1}{2}\left [ \frac{n(n+1)(2n+1)}{6} \right ]$ $\displaystyle +\frac{1}{6}\left [ \frac{n(n+1)}{2} \right ]$
$\displaystyle =\frac{n(n+1)}{12}$ $\displaystyle \left [ n(n+1)+(2n+1)+1 \right ]$
$\displaystyle =\frac{n(n+1)}{12}\left [ n^{2}+n+2n+1+1 \right ]$
$\displaystyle =\frac{n(n+1)}{12}\left [ n^{2}+3n+2 \right ]$
$\displaystyle =\frac{n(n+1)}{12}\left [ (n+1)(n+2) \right ]$
$\displaystyle =\frac{n(n+1)^{2}(n+2)}{12}$
Find the sum to n terms of the following series whose $\displaystyle n^{th}$ term is given by
8. $\displaystyle n(n+1)(n+4)$
$\displaystyle a_{n}=n(n+1)(n+4)$
$\displaystyle =n^{3}+5n^{2}+4n$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}$ $\displaystyle =\sum_{k=1}^{n}(k^{3}+5k^{2}+4k)$
$\displaystyle =\sum_{k=1}^{n}k^{3}+5\sum_{k=1}^{n}k^{2}+4\sum_{k=1}^{n}k$
$\displaystyle =\frac{(n(n+1))^{2}}{4}$ $\displaystyle +5\left [ \frac{n(n+1)(2n+1)}{6} \right ]$ $\displaystyle +4\left [ \frac{n(n+1)}{2} \right ]$
$\displaystyle =\frac{n(n+1)}{2}$ $\displaystyle \left [ \frac{n(n+1)}{2}+\frac{5(2n+1)}{3}+4 \right ]$
$\displaystyle =\frac{n(n+1)}{2}$ $\displaystyle \left [ \frac{3n(n+1)+10(2n+1)+24}{6} \right ]$
$\displaystyle =\frac{n(n+1)}{2}$ $\displaystyle \left [ \frac{3n^{2}+3n+20n+10+24}{6} \right ]$
$\displaystyle =\frac{n(n+1)}{2}\left [ \frac{3n^{2}+23n+34}{6} \right ]$
$\displaystyle =\frac{n(n+1)(3n^{2}+23n+34)}{12}$
9. $\displaystyle n^{2}+2^{n}$
$\displaystyle a_{n}=n^{2}+2^{n}$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}(k^{2}+2^{k})$
$\displaystyle =\sum_{k=1}^{n}k^{2}+\sum_{k=1}^{n}2^{k}$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}$ $\displaystyle +(2^{1}+2^{2}+2^{3}+...+2^{n})$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}+\frac{2(2^{n}-1)}{2-1}$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}+2(2^{n}-1)$
10. $\displaystyle (2n-1)^{2}$
$\displaystyle a_{n}=(2n-1)^{2}$
$\displaystyle a_{n}=4n^{2}-4n+1$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}$ $\displaystyle=\sum_{k=1}^{n}(4k^{2}-4k+1)$
$\displaystyle=4\sum_{k=1}^{n}k^{2}-4\sum_{k=1}^{n}k+\sum_{k=1}^{n}1$
$\displaystyle =4\left [ \frac{n(n+1)(2n+1)}{6} \right ]-4\left [ \frac{n(n+1)}{2} \right ]+n$
$\displaystyle =n\left [ \frac{2(n+1)(2n+1)}{3} -2(n+1)+1\right ]$
$\displaystyle =n\left [ \frac{4n^{2}+6n+2}{3} -2(n+1)+1\right ]$
$\displaystyle =n\left [ \frac{4n^{2}+6n+2-6n-6+3}{3}\right ]$
$\displaystyle =n\left [ \frac{4n^{2}-1}{3}\right ]$
$\displaystyle = \frac{n(2n+1)(2n-1)}{3}$
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