Question

I really really need help :'( I'm about to cry I'm so frustrated! Algebra 2 help please! 5∑n=1 (3n+1). I need the number of terms, the first, last, and evaluation of the sum :(

Answer 1

is the question

\[\sum_{n=1}^{5} 3n + 1\]

Answer 2

I need the first term, the last term, evaluation of the sum and the number of terms :(

Answer 3

Yes it is

Answer 4

ok... so all you do is substitute

\[n = 1 .... 3 \times 1 + 1 = 4\]

\[n = 2 .... 3\times 2 + 1 = 7\]

\[n = 3.... 3 \times 3 + 1 = 10\]

\[n = 4 .... 3 \times 4 + 1 = 13\]

\[n = 5 .... 3 \times 5 + 1 = 16\]

then just add the answers 4 + 7 + 10 + 13 + 16 =

hope this helps

Answer 5

so if you need 1st term a = 4

last term l = 16

the sum can be found using

\[S_{5} = \frac{5}{2} \times( a + l) ..... =.... \frac{5}{2} \times ( 4 + 16)\]

this is the more algebraic approach.

Answer 6

Can you tell me how the first term is 4?

Answer 7

And also how do you know when you get to the last term, do you just go to 5?

Answer 8

ok... so you have an equation that is

\[3 \times n + 1 \]

substitute n = 1 which is the 1st term in the series... since the values of n start at 1 bottom of the sigma sign and finish at 5... top of the sigma sign.

so substituting you get

\[3 \times 1 + 1 = 3 + 1 = 4\]

hope this helps.

Answer 9

Because I'm working on my next one, 8∑n=1 (2n/3) and I'm getting my first term as .66666666666667

Answer 10

Is that correct?

Answer 11

And thank you so much for your help, it was GREATLY appreciated!!!!!!!!!!!

Answer 12

And my last term came out as 5.33333?

Answer 13

ok... so this has 8 terms

so to find the terms substitute n = 1

n = 1 2/3
n = 2 4/3
n=3 6/3
.
.
.
.

n = 8 16/6

Answer 14

oops should be

n = 8 16/3

I'd leave all the terms as fractions... its just eaiser to add

so the sum = 8/2( 2/3 + 16/3)

4(18/3)

Sum = 24

Answer 15

Thank you so much, I completely understand, really I can not thank you enough :)