Asked by sidra jawad on May 30, 2024
Verified
Write the first five terms of the sequence an=(n+6) !(n+3) !a _ { n } = \frac { ( n + 6 ) ! } { ( n + 3 ) ! }an=(n+3) !(n+6) ! . Assume that n begins with 1.
A) a1=120,a2=210,a3=336,a4=504,a5=720a _ { 1 } = 120 , a _ { 2 } = 210 , a _ { 3 } = 336 , a _ { 4 } = 504 , a _ { 5 } = 720a1=120,a2=210,a3=336,a4=504,a5=720
B) a1=202,a2=330,a3=502,a4=719,a5=990a _ { 1 } = 202 , a _ { 2 } = 330 , a _ { 3 } = 502 , a _ { 4 } = 719 , a _ { 5 } = 990a1=202,a2=330,a3=502,a4=719,a5=990
C) a1=210,a2=336,a3=504,a4=720,a5=990a _ { 1 } = 210 , a _ { 2 } = 336 , a _ { 3 } = 504 , a _ { 4 } = 720 , a _ { 5 } = 990a1=210,a2=336,a3=504,a4=720,a5=990
D) a1=103,a2=194,a3=328,a4=501,a5=719a _ { 1 } = 103 , a _ { 2 } = 194 , a _ { 3 } = 328 , a _ { 4 } = 501 , a _ { 5 } = 719a1=103,a2=194,a3=328,a4=501,a5=719
E) a1=336,a2=504,a3=720,a4=990,a5=1320a _ { 1 } = 336 , a _ { 2 } = 504 , a _ { 3 } = 720 , a _ { 4 } = 990 , a _ { 5 } = 1320a1=336,a2=504,a3=720,a4=990,a5=1320
\(( n + 6 ) !\)
A notation indicating the factorial of n plus 6, meaning the product of all positive integers up to n+6.
\(( n + 3 ) !\)
Denotes the factorial of the sum of a variable n and 3, representing the product of all positive integers up to and including (n+3).
- Identify the beginning five numbers in a prescribed sequence.
- Simplify expressions involving factorials.
Verified Answer
JD
Justyna DrozdJun 03, 2024
Final Answer :
C
Explanation :
Plugging in n = 1 to the formula gives $a_1 = \frac{7!}{4!} = 210$. Similarly, $a_2 = \frac{8!}{5!} = 336$, $a_3 = \frac{9!}{6!} = 504$, $a_4 = \frac{10!}{7!} = 720$, and $a_5 = \frac{11!}{8!} = 990$. Thus, the answer is $\boxed{\textbf{(C)}}$.
Learning Objectives
- Identify the beginning five numbers in a prescribed sequence.
- Simplify expressions involving factorials.