Question

evaluate 12 sigma n=1 2n+5

Answer 1

Problem 1: Evaluate the summation of 3 times negative 2 to the n minus 1 power, ... The one I just posted looks like this: \sum_{n=1}^{12} 2n+5.

Answer 2

Answer:

The sum after evaluation is:

                            216

Step-by-step explanation:

We are asked to evaluate the expression:

$\sum_{n=1}^{12} 2n+5$

Now, this terms could also be expanded by the form:

$\sum_{n=1}^{12}2n+\sum_{n=1}^{12} 5$

i.e. it could also be written as:

$2\cdot \sum_{n=1}^{12}n+5\cdot \sum_{n=1}^{12}1$

We know that:

$\sum_{n=1}^{\infty} n=\dfrac{n(n+1)}{2}$

Also,

$\sum_{k=1}^{n}1=n$

Hence, we get:

$\sum_{n=1}^{12} (2n+5)=2\cdot (\dfrac{12\cdot (12+1)}{2})+5\cdot 12$

i.e.

$\sum_{n=1}^{12} (2n+5)=12\cdot 13+60$

i.e.

$\sum_{n=1}^{12} (2n+5)=156+60$

i.e.

$\sum_{n=1}^{12} (2n+5)=216$