Write the first four terms of the sequence whose general term is given.
-$\left\{ \frac { 3 ^ { n } } { ( n + 3 ) ! } \right\}$
A) $\frac { 1 } { 8 } , \frac { 3 } { 40 } , \frac { 3 } { 80 } , \frac { 9 } { 560 }$

B) $\frac { 1 } { 8 } , \frac { 3 } { 40 } , \frac { 3 } { 40 } , \frac { 9 } { 280 }$

C) $\frac { 3 } { 7 } , \frac { 9 } { 8 } , 3 , \frac { 81 } { 10 }$

D) $\frac { 3 } { 4 } , \frac { 9 } { 5 } , \frac { 9 } { 2 } , \frac { 81 } { 7 }$

Question

Write the first four terms of the sequence whose general term is given.
-$\left\{ \frac { 3 ^ { n } } { ( n + 3 ) ! } \right\}$
A) $\frac { 1 } { 8 } , \frac { 3 } { 40 } , \frac { 3 } { 80 } , \frac { 9 } { 560 }$

B) $\frac { 1 } { 8 } , \frac { 3 } { 40 } , \frac { 3 } { 40 } , \frac { 9 } { 280 }$

C) $\frac { 3 } { 7 } , \frac { 9 } { 8 } , 3 , \frac { 81 } { 10 }$

D) $\frac { 3 } { 4 } , \frac { 9 } { 5 } , \frac { 9 } { 2 } , \frac { 81 } { 7 }$

Quizplus · Accepted Answer

To find the first four terms of the sequence, substitute $n = 1, 2, 3, 4$ into the general term $\frac{3^n}{(n+3)!}$. For $n=1$, the term is $\frac{3^1}{4!} = \frac{3}{24} = \frac{1}{8}$. For $n=2$, the term is $\frac{3^2}{5!} = \frac{9}{120} = \frac{3}{40}$. For $n=3$, the term is $\frac{3^3}{6!} = \frac{27}{720} = \frac{3}{80}$. For $n=4$, the term is $\frac{3^4}{7!} = \frac{81}{5040} = \frac{9}{560}$.

Write the First Four Terms of the Sequence Whose General {3n(n+3)!}\left\{ \frac { 3 ^ { n } } { ( n + 3 ) ! } \right\}{(n+3)!3n​}

Write the First Four Terms of the Sequence Whose General $\left\{ \frac { 3 ^ { n } } { ( n + 3 ) ! } \right\}$