# Question #ef517

Dec 10, 2017

Because summations are linear we can move the 6 outside:

${\sum}_{i = 7}^{26} 6 i = 6 {\sum}_{i = 7}^{26} i$

The summation from 7 to 26 is the same as the summation from 1 to 26 minus the summation from 1 to 6:

${\sum}_{i = 7}^{26} 6 i = 6 \left({\sum}_{i = 1}^{26} i - {\sum}_{i = 1}^{6} i\right)$

We know that the summation of 1 to n is $\frac{n \left(n + 1\right)}{2}$

${\sum}_{i = 7}^{26} 6 i = 6 \left(\frac{26 \left(27\right)}{2} - \frac{6 \left(7\right)}{2}\right)$

${\sum}_{i = 7}^{26} 6 i = 6 \left(330\right)$

${\sum}_{i = 7}^{26} 6 i = 1980$