Hello, Since I am having some difficulties with Geometric Series problem, I thought It was kind of nice to post some problems, some which are very challenging, to those interested. The answers I did not post, but if anyone wants them, they can try the questions and are invited to ask me for the answers. Feel free and try these problem, post comments with your answers!!
Ready?
There are Sequence and Series problems! GOOD LUCK!
Find all of the terms of the finite sequence?
1a.
; 1 < n < 5
Find the first five terms and the twelfth term of the infinite sequence.
2a.
Write a formula for the nth term of given the infinite sequence.
3a.
1. Write a formula for the nth term of the geometric sequence 7, 28, 112, 448, .... Do not use a recursive formula.
2. Write a formula for the nth term of the geometric sequence 16, - 4, 1, -1/4, .... Do not use a recursive formula.
3. Find the first term of a geometric sequence with a fifth term of 32 and a common ratio of -2.
4. Find the common ratio for a geometric sequence with a first term of 3/4 and a third term of 27/16.
5. Find the sum of the finite geometric series 3 - 6 + 12 - 24 + 48 - 96.
6. Write a formula for the nth term of the given geometric sequence. Do not use a recursive formula.
125, 25, 5, 1, ...
4, -12, 36, -108, ...
7. Find the sum of the given finite geometric series.
2 + 14 + 98 + 686 + 4802 + 33614 + 235298
2 comments:
Melissa my answer to the following questions are:
1a. :
-1/8, 1/16, -1/32, 1/64, -1/128
2a. :
first five terms: plugging in the numbers one to five in the equation I get the following answers:
1, e, e^2/2, e^3/6, e^4/24
to find the twelve term you plug in number twelve in the eaquation and the answer is:
e^11/39916800
Problem 3a
e, -e^4, e^9, -e^16
Find a formula for the nth term
First of all you identify that the exponent over each term in n^2. Once you notice this you can find that excluding the negatives, the formula would be e^(n^2).
The problem is to make the formula give one term negative and the next once positive.
So first we analize that when n is even tn is negative and when n is odd then tn is positive.
So we start with -1 and raise it to n+1, remember that when you raise a negative number to an odd exponent the answer will still be negative.
therefore by raising -1 to the 3 power, we get a odd number, what we need for the even terms.
So by adding n 1 more we get the opposite sign. This way we accomplish our purpuse.
So the final formula is Tn= ((-1)^(n+1))(e^(n^2))
Post a Comment