Question

The sum of n terms of a series is given by the expression,
\[S_n = 3(-2)^n+5\]

Determine the 3rd term in the series.

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If I'm correct, the first term can be found by inserting n=1 as the sum of the first term. That gives me -1. Then I found the sum of two terms using the above formula, giving me -1.

Then,
\[a+ar=-1\]
\[(-1)+(-1)(\frac{-9}{8})\]
to find r. Yet it doesn't give me the right answer, as I found three individual terms, added them up, and they give a different number than the given formula.

Answer 1

the third partial sum? or the third term of the sequence?

Answer 2

and does n start at 0 or 1? different texts have different ideas for that as well

Answer 3

Third term of the sequence. As for n, it does not specify. I tried both n=0 and n=1 to no avail.

Answer 4

im wondering then if its talking about the initial terms that are being summed up

Answer 5

s1 = a1 by default for the first term

Answer 6

s2 = s1 + a2

a2 = s2 - s1

Answer 7

s3 = s1 + s2 - s1 + a3

Answer 8

a3 = s3 - s2 is what it might turn out to be

Answer 9

does that make sense?

Answer 10

Going by your last equation, I believe you are subtracting adjacent sums so it leaves one 'left over' which would be the nth term. I believe I have tried this before, but I will try it again.

Answer 11

since youve tried the partial sum approach to no avail, this is the only thing that would make sense for me

Answer 12

Hmm, tried both n=0 and n=1 for the first term and then used ar^2 or ar = the number I got that is supposedly the term to find a, and then working forwards, my sums are still different.

Answer 13

i get -36 for an answer

a1 = -1
a2 = 18
a3 = -36
---------
s3 = -19

but s3 = 3(-2)^3 + 5 = -21 hmmm

Answer 14

Same here... I tried a=8 (n=0) as well to no avail.

Answer 15

-1 + a2 + a3 = -21 , assuming n=1,2,3,...

s2 = -1 + a2 = 3(-2)^2+5 = 3(4)+5 = 17

-1 + a2 = 17, a2 = 18

-1+18 + a3 = -21

a3 = -21+1-18 = -38

Answer 16

so my mathing was off the first time, but the idea was sound enough lol

Answer 17

you might wanna check that for n starting at 0 just in case tho

Answer 18

in which case a3 = a2 = 18

Answer 19

Yes, -38 seems to work. I'm still trying to figure out why my original method of ar = 17 didn't work though.

Answer 20

probably due to some inherent flaw in the concept ;)

Answer 21

Do you know the flaw? I kinda think I need to use two equations to solve it, since one doesn't really prove anything except for that case.

Answer 22

you say the sum of 1 term is -1

then you say the sum of the first 2 terms is ... -1 again for some reason

Answer 23

and it looks like your trying to fit this into a geometric sequence, which itself could be a mistake to begin with

Answer 24

Oops I meant a + ar=17. I just assumed it was a geometric sequence because that's what I'm studying right now...

Anyways, thanks for showing me your method. It makes sense and it works, so I'm no complaining :)

Answer 25

youre welcome, and good luck :)

Answer 26

Wait, doesn't S3 = 3(-2)^3+5 = -19, not -21?

Answer 27

\(S_3=3(-2)^3+5=3(-8)+5=-24+5=-19\)

So @amistre64's first attempt was correct I think. I am also getting \(a_3=-36\).

Answer 28

@KingGeorge - Which approach did you use? Subtracting S3-S2?

Answer 29

\(S_1=-1\), so \(a_1=-1\)
\(S_2=17\) so \(a_2=S_2-a_1=17-(-1)=18\)
\(S_3=-19\) so \(a_3=S_3-(a_1+a_2)=-19 -(18-1)=-19-17=-36\)

Answer 30

In short, \(S_3-S_2\)

Answer 31

Okay, thank you!

Answer 32

You're welcome.