Menggunakan sifat operasi sumasi

1. $\sum_{i=1}^nc=cn$
2. $\sum_{i=1}^nca_i=c\sum_{i=1}^na_i$
3. $\sum_{i=1}^n\left(a_i\pm b_i\right)=\sum_{i=1}^na_i\pm\sum_{i=1}^nb_i$
4. $\sum_{i=m+1}^na_i=\sum_{i=1}^na_i-\sum_{i=1}^ma_i$
5. $\sum_{i=m}^na_i=\sum_{i=m+p}^{n+p}a_{i-p}$

dengan menggunakan sifat 1,2,3,4,5 maka

$\sum_{i=1}^8\left(2k^2+8k+10\right)=\sum_{i=3}^{10}\left(2\left(k-2\right)^2+8\left(k-2\right)+10\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{i=3}^{10}\left(2\left(k^{2\ }-4k+4\right)+8k-16+10\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{i=3}^{10}\left(2k^{2\ }-8k+8+8k-16+10\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{i=3}^{10}\left(2k^{2\ }-2\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\sum_{i=3}^{10}k^2-\sum_{i=3}^{10}2$

$$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\sum_{i=3}^{10}k^2-\left(\sum_{i=1}^{10}2-\sum_{i=1}^22\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\sum_{i=3}^{10}k^2-\left(20-4\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\sum_{i=3}^{10}k^2-16$

Jadi, bentuk yang ekuivalen dengan notasi sigma tersebut adalah

$2\sum_{i=3}^{10}k^2-16\$