7.06

1. find the first six terms of the sequence.
a1 = -7, an = 4 • an-1

a) -7, -28, -112, -448, -1792, -7168
b) -28, -112, -448, -1792, -7168, -28,672
c) -7, -28, -24, -20, -16, -12
d) 0, 4, -28, -24, -20, -16

2. find an equation for the nth term of the arithmetic sequence.
-13, -8, -3, 2,

an = -13 x 5(n - 1)
an = -13 + 5(n - 1)
an = -13 + 5(n + 2)
an = -13 + 5(n + 1)

3. find an equation for the nth term of the arithmetic sequence.
a15 = -53, a16 = -5

a) an = -725 - 48(n - 1)
b) an = -725 + 48(n + 1)
c) an = -725 + 48(n - 1)
d) an = -725 - 48(n + 1)

4. determine whether the sequence converges or diverges. if it converges, give the limit.
11, 44, 176, 704,

a) diverges
b) converges; 231
c) converges; 3751
d) converges; 935

5. find an equation for the nth term of the sequence.
-4, -16, -64, -256,

a) an = 4 • -4n
b) an = 4 • -4n + 1
c) an = -4 • 4n
d) an = -4 • 4n - 1

6. find an equation for the nth term of a geometric sequence where the second and fifth terms are -2 and 16, respectively.

a) an = 1 • (-2)n - 1
b) an = 1 • 2n
c) an = 1 • (-2)n + 1
d) an = 1 • 2n - 1

7. write the sum using summation notation, assuming the suggested pattern continues.
4 - 24 + 144 - 864 +

a) summation of four times six to the power of n from n equals zero to infinity
b) summation of four times negative six to the power of n from n equals zero to infinity
c) summation of four times negative six to the power of the quantity n minus one from n equals zero to infinity
d) summation of four times six to the power of the quantity n plus one from n equals zero to infinity

8. write the sum using summation notation, assuming the suggested pattern continues.
-3 + 6 + 15 + 24 + + 132

a) summation of negative 27 times n from n equals 0 to infinity
b) summation of negative 27 times n from n equals 0 to 15
c) summation of the quantity negative 3 plus 9 n from n equals 0 to infinity
d) summation of the quantity negative 3 plus 9 n from n equals 0 to 15

9. write the sum using summation notation, assuming the suggested pattern continues.
343 + 512 + 729 + 1000 + + n3

a) summation of the quantity n minus 1 cubed from n equals 7 to infinity
b) summation of n cubed from n equals 7 to infinity
c) summation of n cubed from n equals 8 to infinity
d) summation of the quantity n plus 1 cubed from n equals 7 to infinity

10. find the sum of the arithmetic sequence.
3, 5, 7, 9, 21

a) 39
b) 120
c) 20
d) 23

11. find the sum of the geometric sequence.
4 divided by 3, 16 divided by 3, 64 divided by 3, 256 divided by 3, 1024 divided by 3

a) 1363 divided by 3
b) 1364 divided by 15
c) 1364 divided by 3
d) 1363 divided by 15

12. an auditorium has 20 rows with 10 seats in the first row, 12 in the second row, 14 in the third row, and so forth. how many seats are in the auditorium?

a) 390
b) 580
c) 620
d) 400

13. use mathematical induction to prove the statement is true for all positive integers n.
10 + 20 + 30 + + 10n = 5n(n + 1)

14. a certain species of tree grows an average of 4.2 cm per week. write an equation for the sequence that represents the weekly height of this tree in centimeters if the measurements begin when the tree is 300 centimeters tall.

1. A. According to the expression a_n=4*a_n-1, each term after a1 is four times the previous term. The first term is -7 as given, 2nd term should be -7*4=-28, 3rd term is -28*4=-112, ... A is the correct answer. 

2. B. The sequence is -13, -8, -3, 2... It's obvious that each term is equal to the previous term plus 5. This is an arithmetic sequence with initial term -13 and common difference 5. We know a1=-13, so a_n=-13+5*(n-1). The answer is B.

3. A. We are given a15=-53, a16=-5. The common difference of the arithmetic sequence is -5-(-53)=48. The formula for a_n term is a1+48*(n-1). We know that a15=-13; plug in n=15, a15=-53=a1+48*(15-1), a1=-725. So a_n=-725+48*(n-1).

4. Diverge. We are given a few terms, 11, 44, 176, 704... Observe that each term is four times the previous one. 11*4=44, 44*4=176, 176*4=704... This is a geometric series with common ratio>1. You can keep multiplying by 4 and the series goes to infinity, so it diverges.

5. D. We have -4, -16, -64, -256... Same as above, each term is four times the previous one. The initial term is a1=-4. The common ratio d=4. So a_n=a1*d^(n-1)=-4*4^(n-1)=-4^n. (D).

6. The answer is A. a2=-2, a5=16. Suppose the common ratio is D. a_n=a1*d^(n-1). a2=a1*d; a5=a1*d^4. Plug in a2 and a5: -2=a1*d, 16=a1*d^4. 16/-2=d^3=-8, d=-2, a1=1. So a_n=1*(-2)^(n-1).

7. B. We are given the sequence 4, -24, 144,... Each term is -6 times the previous one. The first term a0=4, the n^th term a_(n-1) is a1*d^n=4*(-6)^n. To express the sum, we simply have to use the sigma notation and sum 4*(-6)^n from n=0 to infinity. The answer is B.

8. D. We are given -3 + 6 + 15 + 24... 132. Each term is equal to the previous one plus 9. First term a0=-3, n^th term a_n-1 is -3+9*n. The last term is 132. 132 =-3+9n, n=15. So we have to sum -3+9n from n=0 to n=15.

9. B. 343 + 512 + 729 + 1000+...  343=7^3, 512=8^3, 729=9^3, 1000=10^3. This is a sequence of perfect cubes. Therefore, the sum is n^3 from n=7 to infinity. (The initial term is 343=7^3).

10. B. We are given 3, 5, 7, 9, ... 21. The common difference is 2. There are (21-3)/2+1=10  terms. The initial term a1=3, and last term is a10=21. The sum is (a1+a10)*10/2=(3+21)*10/2=120.

11. C. 4/3, 16/3, 64/3, 256/3, 1024/3.  Each term is four times the previous one. This is a geometric series with initial term a1=4/3 and common ratio r=4. 1024/3 is the 5th term of the sequence. So sum=a1*(1-r^n)/(1-r)=4/3*(1-4^5)/(1-4)=-4/9*-1023=1364/3.

12. B. 10,12,14,... This is an arithmetic sequence. a1=10, and common difference d=2. There are 20 terms (20 rows). a20=a1+d*(n-1)=10+2*(20-1)=48. So the sum S=(a1+an)*n/2=(10+48)*20/2=580.

13. 10 + 20 + 30 + ... + 10n = 5n(n + 1). When n=1, this expression is true, since 10=5*1*(1+1). Suppose when n=k, this statement is true, then when n=k+1, the left side is 10+...+10n+10(n+1), the right side is 5(n+1)(n+2). The left side adds 10(n+1) compared to the previous one. The right side adds 5(n+1)(n+2)-5n(n+1)=5(n+1)(n+2-n)=10(n+1). So the statement holds true.

14. The height at week 0 is a0=300 (initial height). Common difference is 4.2 (weekly increment). a_n=300+4.2n. At week n, the height of the tree is 300+4.2*n centimeters.

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