Algebra
posted by Jon .
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 = a1/1r
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/(7k)!k!^m7kpk
7/(75)!5!^m75p5
7*6*5*4*3/5*4*3*2*1^m2p5
2520/120
21m^2p^5
3)How many fourdigit 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
1x9x8x7 = 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 "1x9x8x7 = 5040 which is choice D "
should say:
10x9x8x7 = 5040 which is choice D