Evaluate 12 sigma n=1 2n+5

Question

asked Mar 28, 2018 112k views

2 Answers

Ben Mares · Answer 1 · 2018-03-30T10:40:32+0000

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.

Driouxg · Answer 2 · 2018-04-01T20:15:16+0000

The sum after evaluation is:

216

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:

$\sum_(n=1)^(\infty) n=(n(n+1))/(2)$

Also,

$\sum_(k=1)^(n)1=n$

Hence, we get:

$\sum_(n=1)^(12) (2n+5)=2\cdot ((12\cdot (12+1))/(2))+5\cdot 12$

i.e.

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

i.e.

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

i.e.

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