Question

Evaluate the summation of 3 n plus 2, from n equals 1 to 14..

Answer 1

\[\sum_{n=1}^{14} 3n + 2\]

Answer 2

Answer options
A) 39

B) 49

C) 340

D) 343

Answer 3

Well first we can break it up using properties of summation

\[\sum_{i}^{n} (c + ai) = \sum_{i = 1}^{n}c + \sum_{i=1}^{n}ai\]

So here, we would have

\[\sum_{n=1}^{14} 3n + \sum_{n=1}^{14} 2\]

Using the properties of summation we know that

\[\sum_{i}^{n}c = cn\]
and
\[\sum_{i}^{n}i = \frac{n(n + 1)}{2}\]

So lets rewrite yours as

\[3\sum_{n=1}^{14} n + \sum_{n=1}^{14} 2\]

Which would be

\[\large 3\frac{14(15)}{2} + 2(14)\]

What does that come out to?

Answer 4

\[\sum_{n=1}^{14}3n+2=\sum_{n=1}^{14}3n+\sum_{n=1}^{14}2=3\sum_{n=1}^{14}n+\sum_{n=1}^{14}2\]
remember the rule \[\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\]

Answer 5

id I tried to answer it but a reaaaaaaaaally long number came up @johnweldon1993

Answer 6

do the algebra that john listed at the end and you should find a useful answer

Answer 7

the number is long

Answer 8

We have

\[\large \frac{3\times 14\times 15}{2} + 2(14)\]

Answer 9

So 3*14*15 = 630

That divided by 2 = 315

And 315 + 2(14)

2 times 14 = 28

315 + 28 = ?

Answer 10

3433

Answer 11

...assuming you accidentally put an additional 3 at the end...

Answer 12

yeah sorry lol