Question

Find, showing all working, a recursive denition of the sequence with general

term

tn = 3 (n + 1)!/2n ; n >= 1

Answer 1

$t_n =\frac{ 3 (n + 1)!}{2n} ; n \geq1$; will be the general term of the sequence.

Let's find the next term:

$t_{n +1}=\frac{ 3 ((n + 1)+1)!}{2(n+1)}=\frac{ 3 (n+2)!}{2(n+1)}=$

$=\frac{ 3 (n+2)(n+1)n\times...\times1}{2(n+1)}=\frac{ 3 (n+2)(n+1)!}{2(n+1)}\frac{n}{n}=$

$=\frac{ 3 (n+1)!}{2n}\frac{n(n+2)}{n+1}=t_n\frac{n(n+2)}{n+1}=\frac{n(n+2)}{n+1}t_n.$

Also we have to find the first term:

$t_1 =\frac{ 3 (1 + 1)!}{2\cdot 1}=\frac{3\cdot 2!}{2}=\frac{3\cdot 2}{2}=3.$

So, the recursive definition of the sequence is as follows:

$t_1=3,$

$t_{n+1}=\frac{n(n+2)}{n+1}t_n.$

Answer:

$t_1=3,$

$t_{n+1}=\frac{n(n+2)}{n+1}t_n.$

But if we have $2^n$ in denominator, the solution is the next:

$t_n =\frac{ 3 (n + 1)!}{2^n} ; n \geq 1$; will be the general term of the sequence.

Let's find the next term:

$t_{n +1}=\frac{ 3 ((n + 1)+1)!}{2^{n+1}}=\frac{ 3 (n+2)!}{2^{n+1}}=$

$=\frac{ 3 (n+2)(n+1)n\times...\times1}{2^n\cdot 2^1}=\frac{ 3(n+1)!}{2^n}\frac{n+2}{2}=t_n\frac{n+2}{2}.$

Also we have to find the first term:

$t_1 =\frac{ 3 (1 + 1)!}{2^1}=\frac{3\cdot 2!}{2}=\frac{3\cdot 2}{2}=3.$

So, the recursive definition of the sequence is as follows:

$t_1=3,$

$t_{n+1}=\frac{n+2}{2}t_n.$

Answer:

$t_1=3,$

$t_{n+1}=\frac{n+2}{2}t_n.$