Write the first five terms of the sequence. (Assume that $n$ begins with 0 .)
$$a _ { n } = \frac { 3 ^ { n } } { ( n + 1 ) ! }$$
A) $1 , \frac { 3 } { 2 } , \frac { 3 } { 2 } , \frac { 9 } { 8 } , \frac { 27 } { 40 }$
B) $\frac { 3 } { 2 } , \frac { 9 } { 8 } , \frac { 27 } { 40 } , \frac { 27 } { 80 } , \frac { 81 } { 560 }$
C) $0 , \frac { 3 } { 2 } , \frac { 9 } { 8 } , \frac { 27 } { 40 } , \frac { 27 } { 80 }$
D) $3 , \frac { 3 } { 8 } , \frac { 9 } { 40 } , \frac { 9 } { 80 } , \frac { 27 } { 560 }$
E) $3 , \frac { 3 } { 2 } , \frac { 3 } { 8 } , \frac { 3 } { 40 } , \frac { 1 } { 80 }$

Question

Quizplus · Accepted Answer

The first five terms of the sequence $a_n = \frac{3^n}{(n+1)!}$ when $n$ starts from 0 are calculated as follows:- For $n=0$, $a_0 = \frac{3^0}{(0+1)!} = \frac{1}{1} = 1$- For $n=1$, $a_1 = \frac{3^1}{(1+1)!} = \frac{3}{2} = \frac{3}{2}$- For $n=2$, $a_2 = \frac{3^2}{(2+1)!} = \frac{9}{6} = \frac{3}{2}$- For $n=3$, $a_3 = \frac{3^3}{(3+1)!} = \frac{27}{24} = \frac{9}{8}$- For $n=4$, $a_4 = \frac{3^4}{(4+1)!} = \frac{81}{120} = \frac{27}{40}$Thus, the first five terms are $1, \frac{3}{2}, \frac{3}{2}, \frac{9}{8}, \frac{27}{40}$.

Write the First Five Terms of the Sequence nnn Begins with 0

Write the First Five Terms of the Sequence $n$ Begins with 0