1)Find the sum of the infinite geometric series: 1 + 3/5 + 9/25 + ..., if it exists.
A)5/3
B)5/2
C)3/5
D)does not exist
I chose B
(3/5)/1 = .6
S = a1/1-r
1/1-.6
2.5
2)Use the Binomial Theorem to find the sixth term in the expansion of (m+2p)^7.
A)21m^2p^5
B)672m^2p^5
C)32m^2p^5
D)448mp^6
7/(7-k)!k!^m7-kpk
7/(7-5)!5!^m7-5p5
7*6*5*4*3/5*4*3*2*1^m2p5
2520/120
21m^2p^5
3)How many four-digit numerical codes can be created if no digit may be repeated?
A)10,000
B)24
C)3024
D)5040
I chose A
10*10*10*10
10,000
#2 You forgot that 2 is also raised to the fifth.
C(7,5)(m^2)(2p(^5
= 21(2^5)m^2p^5
=672m^2p^5
#3. Digits may NOT be repeated, and probably cannot start with a zero.
so 9x9x8x7 = 4536, which is none of their choices.
if zeros would be allowed in the first position, then the answer would be
1-x9x8x7 = 5040 which is choice D
your choice A would include zeros, and repeating digits, even 0000,and 6666.
I would guess that D is the choice they are after.
#1 is ok
typo in "1-x9x8x7 = 5040 which is choice D "
should say:
10x9x8x7 = 5040 which is choice D
Your answers are correct for questions 1, 2, and 3.
1) The sum of the infinite geometric series 1 + 3/5 + 9/25 + ... is equal to S = a1/(1-r), which in this case is 1/(1-3/5) = 5/2.
2) Using the binomial theorem, the sixth term in the expansion of (m+2p)^7 is given by (7 choose 5) * (m^(7-5)) * (2p)^5 = 21 * m^2 * p^5.
3) The number of four-digit numerical codes that can be created without repeating digits is equal to 10 choices for the first digit, 9 choices for the second digit (since one digit has already been used), 8 choices for the third digit, and 7 choices for the fourth digit. So, the total number of codes is 10 * 9 * 8 * 7 = 5040.
Therefore, the correct answers are B, A, and D for questions 1, 2, and 3, respectively.
For the first question, to find the sum of an infinite geometric series, we need to determine if the series converges. In this case, the common ratio is 3/5. If the absolute value of the common ratio is less than 1, the series converges. Here, the absolute value of 3/5 is less than 1, so the series converges.
To find the sum of the infinite geometric series, we can use the formula S = a1/(1 - r), where a1 is the first term and r is the common ratio. In this case, a1 is 1 and r is 3/5.
Plugging the values into the formula, we get S = 1/(1 - 3/5) = 1/(2/5) = 5/2.
Therefore, the answer is B) 5/2.
For the second question, to find the 6th term in the expansion of (m+2p)^7 using the Binomial Theorem, we can use the formula: (n choose k) * a^(n-k) * b^k, where n is the power of the binomial, k is the term number, a is the first term, and b is the second term.
In this case, (m+2p)^7, n is 7, a is m, and b is 2p.
Plugging the values into the formula, we get (7 choose 6) * m^(7-6) * (2p)^6 = 7 * m^1 * 64p^6 = 448mp^6.
Therefore, the answer is D) 448mp^6.
For the third question, to calculate the number of four-digit numerical codes without repeating digits, we need to count the possible combinations for each digit. Since no digit can be repeated, we have 10 choices for the first digit, 9 choices for the second digit, 8 choices for the third digit, and 7 choices for the fourth digit.
Using the multiplication principle, we multiply the number of choices for each digit to get the total number of combinations: 10 * 9 * 8 * 7 = 5,040.
Therefore, the answer is D) 5,040.