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Rama09 [41]
3 years ago
9

Evaluate 12 sigma n=1 2n+5

Mathematics
2 answers:
True [87]3 years ago
5 0
Problem 1: Evaluate<span> the summation of 3 times negative 2 to the n minus 1 power, ... The one I just posted looks like this: \sum_{</span>n=1}^{12<span>} </span>2n+5<span>.</span>
Radda [10]3 years ago
3 0
<h2>Answer:</h2>

The sum after evaluation is:

                            216

<h2>Step-by-step explanation:</h2>

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

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Answer:

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Step-by-step explanation:

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