Question

Does anyone know how to write this in sigma notation?

-2, + 1 + 4 + ... + 13


Please explain(:

Answer 1

So its -2+(-2+3*1)+(-2+3*2)+(-2+3*3)+...an

Answer 2

so adding 3... Where do we go from there?(:

Answer 3

Can you come up with an expression for the n'th term of the sequence that is being added here.
-2,(-2+3*1),(-2+3*2),(-2+3*3),...an

Answer 4

plug in for this formula tn= t1 +(n-1)d ?????

Answer 5

Yes

Answer 6

tn = -2 + (n-1) 3

so -2 + 3n - 3

-5 + 3n

What do we do next to put it in sigma?

Answer 7

Over what values of n are you adding. What is the n for the first value and for the last value?

Answer 8

13?

Answer 9

No what value of n gives you 13 from your tn equation. Also which one gives you -2?

Answer 10

im not sure

Answer 11

You have tn=-5 + 3n. Now set tn=-2 and solve for n.

Answer 12

\[\Large \sum_{n=0}^{5} 3n-2\]

Answer 13

wait why do we set tn = -2

Answer 14

help(:

Answer 15

geerky that isn't the right answer...