Question

Evaluate the summation of 2 n plus 5, from n equals 1 to 12..

A. 29
B. 36
C. 216
D. 432

Answer 1

the correct answer is C)216 / Let f(n) = 2n + 5. Then we must simply calculate (f1), f(2), ..., (f12). And we have that f(1) = 2*1 + 5 <=> f(1) = 7; f(2) = 2*2 + 5 <=> f(2) = 9; f(3) = 2*3 + 5 <=> f(3) = 11. And so on. It's easy to see that the rest of the results will be 13, 15, 17, 19, 21 and so on. So the summation will be S = 7 + 9 + 11 +... + 29 <=> S = 216

Answer 2

Option: C is the correct answer.

C. 216

Explanation:

We are asked to evaluate the sum of the following summation:

`\sum_(n=1)^(12) (2n+5)`

Now, this summation could also be written as:

`\sum_(n=1)^(12) (2n+5)=\sum_(n=1)^(12) 2n+\sum_(n=1)^(12) 5`

i.e.

`\sum_(n=1)^(12) (2n+5)=2\sum_(n=1)^(12) n+\sum_(n=1)^(12) 5`

Now, we know that:

`\sum_(n=1)^(12) n=1+2+3+4+....+12\\\\i.e.\\\\\sum_(n=1)^(12) n=(12(12+1))/(2)`

Since,

`\sum_(i=1)^(n) i=(n(n+1))/(2)`

i.e.

`\sum_(n=1)^(12) n=78`

and

`\sum_(n=1)^(12) 5=5+5+....+5\ (\text{twelve\ times})\\\\i.e.\\\\\sum_(n=1)^(12) 12=12* 5\\\\i.e.\\\\\sum_(n=1)^(12) n=60`

Hence, we have:

`\sum_(n=1)^(12) (2n+5)=2* 78+60`

i.e.

`\sum_(n=1)^(12) (2n+5)=156+60`

i.e.

`\sum_(n=1)^(12) (2n+5)=216`