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

A. 75
B. 77
C. 96
D. 98

User Keith K
by
7.2k points

2 Answers

2 votes

Answer:

A). 75

Explanation:

User Hadi Aghandeh
by
7.0k points
4 votes


\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

User Vijikumar M
by
6.3k points
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