Question

Please help. For the sequence an=3(1/2)^(n-1), S1=3 and S2=4.5. How would you find S3, S4, S5, S6, and S7? (Carry out as many decimal places as calculator will allow.)

Answer 1

Do you know what type of sequence this is? (arithmetic vs. geometric)

Answer 2

The sequence looks like this:
\[a_n = 3(\frac{1}{2})^{n-1}\]
\[3, \frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \ldots \]

Answer 3

It's a geometric sequence. Thank you for your help :)

Answer 4

nice :) alright, since we know its a geometric sequence, there is actually a formula for the sum of the first n terms. it is:
\[\S_n = \frac{a_0(r^n-1)}{r-1}\]
Where r is the number you are multiplying each term by to get the next term, a0 is the first term, and n is the number of terms you want in the sum.

Answer 5

So, if we wanted say S3, we would do:
\[a_0 = 3, r = \frac{1}{2}, n = 3\]
\[\S_3 =\frac{3(\frac{1}{2}^3-1)}{1-\frac{1}{2}}\]
every time you want the sum for more terms, just change n. ao and r will always be the same.

Answer 6

Oh o.k. I get it now. :D

Answer 7

in the post above it should be 1/2 - 1 int the denominator of that fraction, my bad >.<

Answer 8

You have to find a general formula for the series which can only be obtained from converging sequences like this one is( i.e. it gets smaller with every term). You then go from term one, to the nth term in the sequence looking for a pattern or a formula in the series.
\[\sum_{1}^{n}3(1/2)^{n-1}\] = 3(1/2)^1-1 + 3(1/2)^(2-1)... 3(1/2)^{n-1}

So if we took lets say, \[\sum_{1}^{n-1}\] from the the original series, we can get a "general" formula for any term we want in the sequence.

eg. \[\sum_{1}^{n} - \sum_{1}^{n-1} = 3(1/2)^{1-1} - 3(1/2)^{1-2}\] .. \[3(1/2)^{n-1} - 3(1/2)^{n-2} + 3(1/2)^{n} - 3(1/2)^{n-1}\] = -6 + 3(1/2)^n

so \[T_{n}\] or the general term will be simplified as, ([1/2^{n} - 2]

Then plug in any number of the term you want like S4, S5 ie 5 for n and you will get your term if i did it right!

Answer 9

I mean 3*(1/2^n - 2) sorry mis-type