menggunakan sifat operasi sumasi

1. $\sum_{i=1}^nc=cn$
2. $\sum_{i=1}^n\left(a_i\pm b_i\right)=\sum_{i=1}^na_i\pm\sum_{i=1}^nb_i$
3. $\sum_{i=1}^nca_i=c\sum_{i=1}^na_{i\ }$
4. jika $m$ bilangan bulat dengan $1<m<n$ dan $a_i$ menyatakan rumus suatu barisan bilangan, maka berlaku $\sum_{i=m+1}^na_i=\sum_{i=1}^na_{i\ }-\sum_{i=1}^ma_i$
5. $\sum_{i=1}^ni=1+2+3+...+n=\frac{n\left(n+1\right)}{2}$

maka

$\sum_{i=4}^{41}\left(4i-3\right)=\sum_{i=4}^{41}4i-\sum_{i=4}^{41}3$ (sifat 2)

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\sum_{i=4}^{41}i-\sum_{i=4}^{41}3$ (sifat 3)

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left(\sum_{i=1}^{41}i-\sum_{i=1}^3i\right)-\left(\sum_{i=1}^{41}3-\sum_{i=1}^33\right)$ (sifat 4)

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left(\frac{41\left(41+1\right)}{2}-\frac{3\left(3+1\right)}{2}\right)-\left(\sum_{i=1}^{41}3-\sum_{i=1}^33\right)$ (sifat 5)

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left(\frac{41\left(41+1\right)}{2}-\frac{3\left(3+1\right)}{2}\right)-\left(3\left(41\right)-3\left(3\right)\right)$ (sifat 1)

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left(\frac{41\left(42\right)}{2}-\frac{3\left(4\right)}{2}\right)-\left(123-9\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\left(1722-12\right)-\left(114\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\left(1710\right)-\left(114\right)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3420-114$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3306$

Jadi, nilai dari $\sum_{i=4}^{41}\left(4i-3\right)=3.306$