Download Practice Solutions for College Algebra II (MATH 010) Spring 2015 Exam IV and more Schemes and Mind Maps Algebra in PDF only on Docsity!
COLLEGE ALGEBRA II (MATH 010) SPRING 2015 PRACTICE FOR EXAM IV SOLUTIONS
- Find the first four terms and the 100th term of each sequence. (a) an = (^) n+1^1 (b) an = (−1)
n n^2 Solution: (a) a 1 = 12 (b) a 1 = −^11 a 2 = 13 a 2 = (^14) a 3 = 14 a 3 = −^19 a 4 = 15 a 4 = 161 a 100 = 1011 a 100 = (^10012)
- Find the first five terms of the recursive sequence. an = 2(an− 1 − 2) and a 1 = 3 Solution: a 1 = 3 a 2 = 2(3 − 2) = 2 a 3 = 2(2 − 2) = 0 a 4 = 2(0 − 2) = − 4 a 5 = 2(− 4 − 2) = − 12
- Find the n-th term of the sequence whose first several terms are given. 1 , 4 , 7 , 10 ,... Solution: a 1 = 1 a 2 = 4 = 1 + 3 a 3 = 7 = 1 + 3 + 3 a 4 = 10 = 1 + 3 + 3 + 3 an = 1 + 3(n − 1) = 3n − 2
- Find the sum. ∑ 3 k= 1 k Solution: ∑ 3 k= 1 k =^
1 1 +^
1 2 +^
1 3 =^
6 6 +^
3 6 +^
2 6 =^
11 6
- Write the sum using sigma notation. 12 + 2^2 + 3^2 +... + 10^2 Solution: ∑ 10 k=1 k 2
- Determine the common difference, the fifth term, the n-th term, and the 100-th term of the arithmetic se- quence. (a) 2, 5 , 8 , 11 ,... (b) − 12 , − 8 , − 4 , 0 ,... Solution: (a) d = 3 (b) d = 4 a 5 = 14 a 5 = 4 an = 2 + 3(n − 1) an = −12 + 4(n − 1) a 100 = 2 + 3(99) = 299 a 100 = −12 + 4(99) = 384
- Find the partial sum Sn of the arithmetic sequence that satisfies the given conditions. a = 4, d = 2, n = 20 Solution: an = 4 + 2(n − 1) a 20 = 4 + 2(19) = 42 Sn = n(a+ 2 a n) S 20 = 20(4+42 2 ) = 460
- Use Pascal’s triangle to expand (x − 1)^5. Solution: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 (x − 1)^5 = x^5 + 5x^4 (−1) + 10x^3 (−1)^2 + 10x^2 (−1)^3 + 5x(−1)^4 + (−1)^5 = x^5 − 5 x^4 + 10x^3 − 10 x^2 + 5x − 1
- Use the Binomial Theorem to expand (x + y)^6. Solution: (x + y)^6 =
( 6 0
) x^6 +
( 6 1
) x^5 y +
( 6 2
) x^4 y^2 +
( 6 3
) x^3 y^3 +
( 6 4
) x^2 y^4 +
( 6 5
) xy^5 +
( 6 6
) y^6 = x^6 + 6x^5 y + 15x^4 y^2 + 20x^3 y^3 + 15x^2 y^4 + 6xy^5 + y^6
- Find the term containing x^9 in the expansion of (x + 3)^14. Solution: Use ∑n k=
(n k
) an−kbk^ with n = 14, a = x, and b = 3. Since 14 − k = 9, k = 5 and the term containing x^9 is
( 14 5
) x^935 = (2002)x^9 (243) = 486, 486 x^9
- Evaluate each expression. (a) P (8, 3) (b) C(8, 3) Solution: (a) P (8, 3) = (^) (8−8!3)! = 8!5! = 8 · 7 · 6 = 336 (b) C(8, 3) = (^) 3!(88!−3)! = (^) 3!5!8! = 83 ··^72 ··^61 = 56
- In how many ways can a race with five runners be completed? (Assume that there is no tie.) Solution: 5 · 4 · 3 · 2 · 1 · = 5! = 120
- In how many different ways can a president, vice president, and secretary be chosen from a class of 15 students? Solution: Order of selection matters (Permutation) P (15, 3) = 15!12! = 15 · 14 · 13 = 2730
- In how many ways can a committee of three members be chosen from a club of 25 members? Solution: Order of selection does not matter (Combination) C(25, 3) = (^) 3!22!25! = 253 ··^242 ·· 123 = 25 · 4 · 23 = 2300
- An experiment consists of tossing a coin twice.
(a) Find the sample space. (b) Find the probability of getting heads exactly two times. (c) Find the probability of getting heads at least one time. (d) Find the probability of getting heads exactly one time. Solution: (a) S = {HH, HT, T H, T T } (b) E = {HH}. P (E) = n n((ES)) = (^14) (c) F = {HH, HT, T H}. P (F ) = n n((FS^ )) = (^34) (d) G = {HT, T H}. P (G) = n n((GS)) = 24 = (^12)
