Question

Evaluate the sum from n equals three to seven of quantity two n plus five.

Answer 1

$\bf \displaystyle\sum_{n=3}^{7}~2n+5\implies \sum_{n=3}^{7}~2n+\sum_{n=3}^{7}~5\implies 2\sum_{n=3}^{7}~n+\sum_{n=3}^{7}~5$

now let's change the lower bounds to n = 1, and then subtract the first two values' sum, that way we'll end up with the sum from n = 3 to n = 7, why doing so?  because that way we can use the special sum form which requires n = 1.

$\bf \displaystyle\left[2\sum_{n=1}^{7}~ n - 2\sum_{n=1}^{2}~n\right]+\left[ \sum_{n=1}^{7}~5 - \sum_{n=1}^{2}~ 5 \right] \\\\\\ \left[ 2\left( \cfrac{7(7+1)}{2} \right)-2\left( \cfrac{2(2+1)}{2} \right) \right]+\left[\cfrac{}{}(7)(5)-(2)(5) \right] \\\\\\ \left[ 2(7(4)- 2(3)) \cfrac{}{}\right]+\left[ 35-10\cfrac{}{} \right]\implies [2(28-6)]+[25] \\[2em] [2(25)]+[25]\implies 50+25\implies 75$