Question

Let a_n=2_n+6 and let S_n=a_1+a_2+a_3…+a_n find a_(n ) and s_n when n=30

Answer 1

Wouldn't a of n =2(30)+6

Answer 2

No that would be only the 30th term of the sequence.

Answer 3

and that is 66 right

Answer 4

so what is Sn?

Answer 5

\[S _{30}= \sum_{1}^{30}2n+6\]

Answer 6

Oh and I didn't see that they asked for a30 yeah that's the correct way to find it.

Answer 7

I'll help you calculate S30 in just a sec.

Answer 8

ok

Answer 9

\[S _{30}=\sum_{1}^{30}2n+6=2\sum_{1}^{30}n+\sum_{1}^{30}6\]
now you need to know that \[\sum_{1}^{k}n=k(k+1)/2\]and\[\sum_{1}^{k}c=kc\]Can you take it from here?

Answer 10

too complex

Answer 11

What particularly do you not understand?

Answer 12

kindly solve for Sn

Answer 13

Do you understand what I did to break up the 2n+6 to two different sums?

Answer 14

No

Answer 15

Ok that just comes from the linearity of summations. Basically sum(x+y)=sumx+sumy and sum(cx)=csum(x).

Answer 16

I should have mentioned in that property earlier that the c was a constant.

Answer 17

Hmm, take Sn=30/2(2a+(30-1)d)
Can it be solved this way and if then what is d?

Answer 18

what is a and d ?

Answer 19

What is that equation?

Answer 20

formula for finding the sum of a1+a2+a3... a of n

Answer 21

http://www.mathsisfun.com/algebra/partial-sums.html
Check out this site. It is where I am getting these identities from.

Answer 22

Sn=n/2(2a+(n-1)d)

Answer 23

ok

Answer 24

I'll write out the steps of the solution step by step while you read. I'll try to show as much work as possible so feel free to ask questions!

Answer 25

\[S _{30}=\sum_{1}^{30}2n+6=\sum_{1}^{30}2n+\sum_{1}^{30}6\] Then we can pull the 2 out of the first term:\[=2\sum_{1}^{30}n+\sum_{1}^{30}6\]Now I will evaluate the two sums based on those identities I posted earlier.\[=2(30)(30+1)/2+6(30)=930+180=1110\]