Question

evaluate the sum (picture)

Answer 1

Is this a calc related question? just wondering.

Answer 2

algebra two sequences/series

Answer 3

Ohhh okay.

Answer 4

can you help me?

Answer 5

I'm working on the problem:) just a minute :)

Answer 6

ok thanks :)

Answer 7

no i have not done that

Answer 8

still need help?

Answer 9

so you have to do tat thiirty times? or can you use that equation i guess even if i didn't lean it yet?

Answer 10

i got 1350 with that formula.. is that right?

Answer 11

Ahh, i found out what i did wrong. Let's redo this.

Answer 12

nope.

Answer 13

Lol i see what i did...

So we are trying to get a pattern here..

a1 = 3(1)-1 = 2
a2 = 3(2)-1 = 5
a3= 3(3)-1 = 8....

If we keep inputting numbers for n all the way to 30, we'll see that the difference in these numbers in 3, we add 3 to each number to get the next one.

This problem is an arithmetic series in that the formula follows:

an = a1 + (n-1)d

d is the difference in numbers in your series, and a1 is the first number you found. n is the total amount of numbers you want to find the sum of.

We're trying to find our sum all the way to 30, so we'll have
\[\large a_{30}= 2 + (30-1)*3\]

3 = difference in between the preceding term and the term before.
2 = the value your function when n=1
n=30 being the total amount of numbers you're trying to find

Answer 14

ok i understand this. but now is there a way to get the sum?

Answer 15

Just evaluate the equation \[\large a_{30}= 2+3(29)\]

Answer 16

ok thank you!

Answer 17

No problem, do you understand how this works now?

Answer 18

Use the following formulas:
\[\sum_{i=1}^ni=\frac{n(n+1)}{2}\\
\sum_{i=1}^n1=n\]
So, your series can be written as
\[\begin{align*}\sum_{n=1}^{30}(3n-1)&=3\sum_{n=1}^{30}n-\sum_{n=1}^{30}1\\
&=3\cdot\frac{30(30+1)}{2}-30\\
&=1365
\end{align*}\]