Amy's Reflection #26

ok i'm just gonna do a whole bunch of examples from chapter 13 to help jog y'all's memory, kk??

Examples:

1. Find the 32nd term in the sequence: 1,4,7,10...
*figure out whether this sequence is arithmetic or geometric.
*arithmetic:because you're adding 3 each time.
*use the arithmetic formula: tn=t1+(n-1)d
*you are looking for "n" in the formula. So you plug in 32 wherever "n" is in the formula.
*So you get t32=1+(32-1)(3)
*that simplifies to =1+(31)(3)
*So t32=94

2. lim (n infinity) sin (1/n)
sin(1/100) = .010
sin(1/1000) = .0010
sin(1/10000) = .00010

lim (n infinity) n+5/n = 1, because the degree is the same, so the coefficients equal 1

3. where the values of x converge
1+(x-2)+(x-2)^2+(x-2)^3+
r=x-2
x-2<1
-1<1
1<3

4.In the arithmetic sequence:3,5,7,9-- find the 28th term.
t28=3+(27)(2)
t28=3+54
t28=57

5. In the geometric sequence: 2,4,8,16-- find the 10th term
t10= 2*2^9
t10= 2*512
t10= 1024

6. Find the sum of the first ten terms of the series:
2 - 6 + 18 - 54 +...
s10 = 2(1 - (-3)^10) / 1 - (-3)
s10 = -29, 524

*(2 being the first number in the problem, -3 being what you multiply each number by
to get the next term)

7. Find the sum of the first 25 terms of the arithmetic series:
11 + 14 + 17 + 20 +...
tn = 11 +(25 - 1)3
tn = 83
s25 = 25(11 +83) / 2
s25 = 1175

*(11 being the first number in the sequence, 3 being the number you add, Plug 25 into the
n-1 formula because your looking for the 25th term)

And for those who have trouble with the problems the involve sigma:

alrighty so say you have this problem...

write the series expanded form.

the limits of summation are 4 and k=1. the index is k. and the summand is 5k.

to expand it, your answer would be 5+10+15+20.

*you have to have 4 numbers in the series becaue thats the number that is the limit of summation. your summand is 5k so you would multiply 1*5, 2*5, 3*5, 4*5 and your answer is 5+10+15+20.

hoped that help...