k2 + 3k + 2 = (k2 + k) + 2 ( __________ )

Question 1 options:k + 5k + 1k + 3k + 2

## Question 2 (5 points)

The following are defined using recursion formulas. Write the first four terms of each sequence. a1 = 7 and an = an-1 + 5 for n ≥ 2

Question 2 options:8, 13, 21, 227, 12, 17, 226, 14, 18, 214, 11, 17, 20

## Question 3 (5 points)

Write the first four terms of the following sequence whose general term is given.an = 3n

+ 2

Question 3 options:4, 6, 10, 146, 9, 12, 155, 8, 11, 147, 8, 12, 15

## Question 4 (5 points)

If 20 people are selected at random, ﬁnd the probability that at least 2 of them have the same birthday.

Question 4 options:≈ 0.31≈ 0.42≈ 0.45≈ 0.41

## Question 5 (5 points)

Saved

Use the Binomial Theorem to expand the following binomial and express the result in simpliﬁed form.(x2 + 2y)4

Question 5 options:x8+8x6y+24x4y2+32x2y3+16y4x8+8x6y+24x4y2+32x2y3+16y4x8+8x6y+20x4y2+30x2y3+15y4x8+8x6y+20x4y2+30x2y3+15y4x8+18x6y+34x4y2+42x2y3+16y4x8+18x6y+34x4y2+42x2y3+16y4x8+8x6y+14x4y2+22x2y3+26y4x8+8x6y+14x4y2+22x2y3+26y4

## Question 6 (5 points)

Consider the statement “2 is a factor of n2 + 3n.”If n = 1, the statement is “2 is a factor of __________.”If n = 2, the statement is “2 is a factor of __________.”If n = 3, the statement is “2 is a factor of __________.”If n = k + 1, the statement before the algebra is simpliﬁed is “2 is a factor of __________.”If n = k + 1, the statement after the algebra is simpliﬁed is “2 is a factor of __________.”

Question 6 options:4; 15; 28; (k + 1)2 + 3(k + 1); k2 + 5k + 84; 15; 28; (k + 1) + 3(k + 1); k2 + 5k + 84; 20; 28; (k + 1)2 + 3(k + 1); k2 + 5k + 74; 20; 28; (k + 1) + 3(k + 1); k2 + 5k + 74; 10; 18; (k + 1)2 + 3(k + 1); k2 + 5k + 44; 10; 18; (k + 1) + 3(k + 1); k2 + 5k + 44; 15; 18; (k + 1)2 + 3(k + 1); k2 + 5k + 64; 15; 18; (k + 1) + 3(k + 1); k2 + 5k + 6

## Question 7 (5 points)

Use the Binomial Theorem to expand the following binomial and express the result in simpliﬁed form.(2×3 – 1)4

Question 7 options:14×12−22×9+14×6−6×3+114×12-22×9+14×6-6×3+116×12−32×9+24×6−8×3+116×12-32×9+24×6-8×3+115×12−16×9+34×6−10×3+115×12-16×9+34×6-10×3+126×12−42×9+34×6−18×3+126×12-42×9+34×6-18×3+1

## Question 8 (5 points)

Write the first four terms of the following sequence whose general term is given.an = 3nQuestion 8 options:3, 9, 27, 814, 10, 23, 915, 9, 17, 314, 10, 22, 41

## Question 9 (5 points)

Use the formula for the sum of the first n terms of a geometric sequence to solve the following.Find the sum of the first 11 terms of the geometric sequence: 3, -6, 12, -24 . . .

Question 9 options:10452108104782049

## Question 10 (5 points)

A club with ten members is to choose three officers—president, vice president, and secretary-treasurer. If each office is to be held by one person and no person can hold more than one office, in how many ways can those offices be filled?

Question 10 options:650 ways720 ways830 ways675 ways

## Question 11 (5 points)

Use the Binomial Theorem to find a polynomial expansion for the following function.f1(x) = (x – 2)4

Question 11 options:f1(x)=x4−5×3+14×2−42x+26f1x=x4-5×3+14×2-42x+26f1(x)=x4−16×3+18×2−22x+18f1x=x4-16×3+18×2-22x+18f1(x)=x4−18×3+24×2−28x+16f1x=x4-18×3+24×2-28x+16f1(x)=x4−8×3+24×2−32x+16f1x=x4-8×3+24×2-32x+16

## Question 12 (5 points)

Write the first four terms of the following sequence whose general term is given.an = (-3)n

Question 12 options:-4, 9, -25, 31-5, 9, -27, 41-2, 8, -17, 81-3, 9, -27, 81

## Question 13 (5 points)

You volunteer to help drive children at a charity event to the zoo, but you can ﬁt only 8 of the 17 children present in your van. How many different groups of 8 children can you drive?

Question 13 options:32,317 groups23,330 groups24,310 groups25,410 groups

## Question 14 (5 points)

Write the first six terms of the following arithmetic sequence.a1 = 5/2, d = – ½

Question 14 options:3/2, 2, 1/2, 1, 1/4, 07/2, 2, 5/2, 1 ,3/2, 05/2, 2, 3/2, 1, 1/2, 09/2, 2, 5/2, 1, 1/2, 0

## Question 15 (5 points)

If two people are selected at random, the probability that they do not have the same birthday (day and month) is 365/365 * 364/365. (Ignore leap years and assume 365 days in a year.)

Question 15 options:The first person can have any birthday in the year. The second person can have all but one birthday.The second person can have any birthday in the year. The first person can have all but one birthday.The first person cannot a birthday in the year. The second person can have all but one birthday.The first person can have any birthday in the year. The second cannot have all but one birthday.

## Question 16 (5 points)

Write the first six terms of the following arithmetic sequence.an = an-1 – 10, a1 = 30

Question 16 options:40, 30, 20, 0, -20, -1060, 40, 30, 0, -15, -1020, 10, 0, 0, -15, -2030, 20, 10, 0, -10, -20

## Question 17 (5 points)

Find the indicated term of the arithmetic sequence with first term, a1, and common difference, d. Find a200 when a1 = -40, d = 5

Question 17 options:865955678895

## Question 18 (5 points)

Write a formula for the general term (the nth term) of each arithmetic sequence. Do not use a recursion formula. Then use the formula for an to ﬁnd a20, the 20th term of the sequence. an = an-1 – 10, a1 = 30

Question 18 options:an = 60 − 10n; a = −260an = 60 – 10n; a = -260an = 70 − 10n; a = −50an = 70 – 10n; a = -50an = 40 − 10n; a = −160an = 40 – 10n; a = -160an = 10 − 10n; a = −70an = 10 – 10n; a = -70

## Question 19 (5 points)

Saved

How large a group is needed to give a 0.5 chance of at least two people having the same birthday?

Question 19 options:13 people23 people47 people28 people

## Question 20 (5 points)

The following are defined using recursion formulas. Write the first four terms of each sequence. a1 = 4 and an = 2an-1 + 3 for n ≥ 2

Question 20 options:4, 15, 35, 4534, 11, 15, 134, 11, 25, 533, 19, 22, 53