Question

2.) For each sum find:
a.) the number of terms, b.) the first term, c.) the last term, and d.) evaluate the sum.
(4 pts each)
A. 5 B. 8
E (3n+1) E (2n/3)
n=1 n=1

Answer 1

\[\sum_{n=1}^{5}\left( 3n+1 \right)\]
no of terms=5
first term =3*1+1=4
last term=3*5+1=16
\[n=5,\sum=\frac{ n }{ 2 }\left( a+l \right)=\frac{ 5 }{ 2 }\left( 4+16 \right)=?\]

Answer 2

50
?

Answer 3

correct.

Answer 4

So then for the next one? I can do the same. Can you help me set that one up?

Answer 5

yes you should proceed in the same way,i will help.

Answer 6

Ok, I'm going to attempt it, one second!

Answer 7

I'm a little confused..
No. of terms= 8

Answer 8

First term is? 2*3+1?

Answer 9

\[\sum_{n=1}^{8}\frac{ 2n }{ 3 }\]
no .of terms n=8
first term \[a=\frac{ 2*1 }{ 3 }=\frac{ 2 }{ 3 }\]
last term \[l=\frac{ 2*8 }{ 3 }=\frac{ 16 }{ 3 }\]
use the above given formula and find sum.

Answer 10

Could you go into slightly more detail. n=8 E= 8/2(2/3+16/3)

Answer 11

So the answer would be around 24?

Answer 12

@surjithayer

Answer 13

correct.