Question 557365: ``write the first six terms of the sequence
2. an=(n+1)^3
4. an=n/n+3
write the next term int he sequence. then write a formula for the nth term.
6.-3,6,-12,24
write the series with summation notation
8. 4+8+12+16
10. 0+3+6+9+12
find the sum of the series (the symbol is sigma)
12. on top of the sigma is 10, bottom is n=4, and on right is n(2n-1)
14. 30 on top, k=1 on bottom, 4 on right
write a rulen for the nth term of the arithmetic sequence
16. 4, 6, 8, 10, 12
18. d=5, a1=13
20. a4=20, a13=65
write a rule for the nth term of the geometric sequence
26. 6, 12, 24, 48
28. r=3, a1=6
30. a2=50, a6=.005
Answer by CubeyThePenguin(3113)   (Show Source):
You can put this solution on YOUR website! 2. a_n = (n + 1)^3
first 6 terms: 2^3, 3^3, 4^3, 5^3, 6^3, 7^3 ----> 8, 27, 64, 125, 216, 343
4. a_n = n/(n+3)
first 6 terms: 1/4, 2/5, 1/2, 4/7, 5/8, 2/3
6. -3, 6, -12, 24 ---> next term = 24(-2) = -48
formula: a_n = (-3)(-2)^(n-1)
8. 4 + 8 + 12 + 16 = sum n from 1 to 4, 4n
10. 0 + 3 + 6 + 9 + 12 = sum n from 0 to 4, 3n
12. n from 4 to 10, n(2n-1)
sum = (4)(7) + (5)(9) + (6)(11) + (7)(13) + (8)(15) + (9)(17) + (10)(19) = 693
14. k from 1 to 30, 4
sum = 4(30) = 120
16. 4, 6, 8, 10, 12
a_n = 4 + 2(n - 1) = 2n + 2
18. d = 5, a_1 = 13 ----> a_n = 13 + 5(n-1) = 5n + 8
20. a_4 = 20, a_(13) = 65 (use system of equations)
a + 3d = 20
a + 12d = 65
Use any method (graphing, elimination, substitution) to solve and get a = 5, d = 5
equation: a_n = 5 + 5(n-1) = 5n
26. 6, 12, 24, 48
rule: a_n = (6)(2)^(n-1)
28. a_1 = 6, r = 3 ----> a_n = (6)(3)^(n-1)
30. a_2 = 50, a_6 = 0.005
ar = 50
ar^5 = 0.005 ----> a = 500, r = 1/10
formula: a_n = (500)(1/10)^(n-1)
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