+2440 To do this problem, evaluate n at the values from the positive integers n=3 to n=7
When n=3:
| \(2n+3\) | Plug in the appropriate value for n, which is 3, in this case. |
| \(2*3+3\) | |
| \(6+3\) | |
| \(9\) | |
Now, let's evaluate it for n=4:
| \(2n+3\) | Plug in n=4 |
| \(2*4+3\) | |
| \(8+3\) | |
| \(11\) | |
n=5:
| \(2n+3\) | Replace all instances of n with 5 |
| \(2*5+3\) | |
| \(10+3\) | |
| \(13\) | |
n=6:
| \(2n+3\) | Plug in 6 for n |
| \(2*6+3\) | |
| \(12+3\) | |
| \(15\) | |
Now, I do not have to evaluate 7 as I can infer, based on the information above, that the next value in the series will be the sum of the last number in the series and 2. Therefore, 2n+3 when n=7, the expression evaluates to 17.
Now, our last step is too add the series together:
| \(9+11+13+15+17\) | Figure out what this evaluates to. Of course, with addition you can add in whatever order you'd like. I'll use this property to my advantage to ease calculation. |
| \(20+13+15+17\) | |
| \(20+30+15\) | |
| \(50+15\) | |
| \(65\) | |
