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Hi;
3) 162 is the b(5), b(8) is 4374
4) The question has a mistake:
4. v1 = 0.75 and vn = (-2)vk-1 for n>1
You probably mean v(k) = -2 v(k-1)...
That was how the question was written but yeah I think that was what it means.
Hi, this is my last algebra lesson and I need to finish this class before May 12th, please help me! I did the questions that I could the best I could. Here is what I really need help with it! I appreciate it!
For questions 3-5, find the first 4 terms and the 8th term of the recursively-defined sequence.
3. b1 = 2 and bk+1 = 3bk, for k>0
My answer was:
b(1) = 2
b(2) = 3*b1 = 3*2 = 6
b(3) = 3*b2 = 3*6 = 3^2*2 = 9*2 = 18
b(4) = 3*b3 = 3*3^2*2 = 3^3*2 = 54
b(8) = 3^4*2 = 81*2 = 162
My teacher said 8th term not corect, but the others are!
4. v1 = 0.75 and vn = (-2)vk-1 for n>1
My answer was:
v(1) = 3/4
v(2) = -2(3/4)
v(3) = (-2)^2(3/4) = 3
v(4) = (-2)^3(3/4) = -6
v(8) = (-2)^6(3/4) = 48
She said: 2nd and 8th term incorrect
5. c1 = 2, c2 = -1, and ck+2 = ck + ck+1 for k>0
My answer was:
c(1) = 2
c(2) = -1
c(3) = 2+-1 = 1
c(4) = -1 + 1 = 0
c(8) = c(6) + c(5) = [c(4)+c(5)] + c(5) = [0+1+1] = 2
8. The third and sixth terms of a geometric sequence are -75 and -9375 respectively. Find the first term, the common ratio, and an explicit rule for the nth term.
She said: #8 is tells you the 3rd term is -75 and the 6th term is -9375. You need to figure out what the 1st term is...and how to get from 1 to the next!
12. Find the sum of the geometric sequence
http://www.sc.whitmoreschool.org/sec/st … age107.gif
For problems 13 and 14, find the sum of the first n terms of the sequence. The sequences are either arithmetic or geometric.
13. -1, 11, -121, ...; n = 9
14. 14, 8, 2,...; n=9
You have read the page I sent you to?
Hi, sorry for the late reply, I was sick. Yes, I read the page and tried the best I can but I couldn't understand or was very confused on what to do.
Okay, it is a bit tricky at first so go here.
http://www.mathsisfun.com/algebra/sigma-notation.html
Try to get something out of that and then if you still can not tackle the problems I will do them.
Hi,
I cannot tackle the problems. They still confuse me.
Hi;
No, they are different. Do you know how to use summation notation?
No I do not know how to use summation notation.
Do not forget to reduce v6.
What help do you need with them?
Just how to do them or if they're the same like number 1 and 2
For problems 9 and 10, write each sum using summation notation.
9.2 + 5 + 8 + 11 + ... + 29
10. 6 - 12 + 24 - 48 + ...
11. Find the sum of the arithmetic sequence 2, 4, 6, 8, ..., 70.
12. Find the sum of the geometric sequence
http://www.sc.whitmoreschool.org/sec/students/classes/scalgebra_2/lesson35_files/clip_image107.gif
For problems 13 and 14, find the sum of the first n terms of the sequence. The sequences are either arithmetic or geometric.
13. -1, 11, -121, ...; n = 9
14. 14, 8, 2,...; n=9
15. Determine whether the infinite series is equal to a real number. If so, find the sum.
http://www.sc.whitmoreschool.org/sec/students/classes/scalgebra_2/lesson35_files/clip_image121.gif
Hi;
Do I stop there or keep going?
You just need v5 and v6 and you are done with the first two questions.
Ah I apologize. I did that and I got:
v5 = 4/5 + 2
v5 = 4/7
v6 = 4/6 + 2
v6 = 4/8
I just need help with the other questions I posted and I'll be good!
90sginger wrote:d100 = 9,500
Should I put this after the d4?
Nope, he was just testing you. If you can do d1, d2, d3, d4, and d100, surely you can do d5 and d6 by yourself.
Ah okay, I did that and I got:
d5 = (5)^2 - 5(5)
d5 = 25 - 25 = 0
d6 = (6)^2 - 5(6)
d6 = 36 - 30 = 6
Do I stop there or keep going?
Yep!
Now what is d100?
d100 = (100)^2 - 5(100)
d100 = 10000 - 500
d100 = 9,500
Should I put this after the d4?
Hmmm. Are you sure 100 * 100 =1000?
100 * 100 = 10,000
That is correct.
What is
for 2)?
d100 = (100)^2 - 5(100)
d100 = 1000 - 500
d100 = 500
Can you reduce that?
v100 = 4/100 + 2 = 4/102 = 2/51
That is correct.
Can you get
v100 = 4/100 + 2 = 4/102
Both v3 and v4 are incorrect, please try again.
v3 = 4/3 + 2 = 4/5
v4 = 4/4 + 2 = 4/6
That is correct and do it for 3).
v3 = 4/3 + 3 = 4/6
v4 = 4/4 + 4 = 4/8
I added v4 just in case.
Start with 1 and plug it into the formula wherever there is an n.
We start with n = 1
What is
V2 = 4/2+2 = 4/4
Should I do the same with V3?
That is correct!
Can you do 1) now or need help.
I need help, if you can!
Hi;
my math is not the greatest
Out of all the billions of people who ever tried to do math only Isaac Newton ever found it easy.
I fixed up your post a bit so everyone can see the problem easily.
Let's try 2)
These are easy, you just start with the counting numbers 1,2,3,4,... and substitute them for n.
Can you do
for me?
d4 = (4)^2 - 5(4)
d4 = 16 - 20
d4 = -4
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