# Thread: Sequence question, is this correct?

1. ## Sequence question, is this correct?

hi everyone

Question

a)Calculate the following sequence:
{(n^3)/(n^3+n^2+1)}^5,n=0

this is how i did it...
n = 1, =1/3
n=2,=8/13
n=3,27/37
n=4,64/81
n=5, 125/151

ans: 1/3,8/13,27/37,64/81,125/151

b){an}^6 ,n=0 if a(n+1)= (an+3)/(an+1), a0=2

a1= (2+3)/(2+1) = 5/3
a2=(3+3)/(3+1)=3/2
a3=(4+3)/(4+1)=7/5
a4=(5+3)/(5+1)=8/6
a5=(6+3)/(6+1)=9/7
a6=(7+3)/(7+1)=10/8

ans:5/3,3/2,7/5,4/3,9/7,5/4

2. Originally Posted by anderson
hi everyone

Question

a)Calculate the following sequence:
{(n^3)/(n^3+n^2+1)}^5,n=0

this is how i did it...
n = 1, =1/3
n=2,=8/13
n=3,27/37
n=4,64/81
n=5, 125/151

ans: 1/3,8/13,27/37,64/81,125/151

b){an}^6 ,n=0 if a(n+1)= (an+3)/(an+1), a0=2

a1= (2+3)/(2+1) = 5/3
a2=(3+3)/(3+1)=3/2
a3=(4+3)/(4+1)=7/5
a4=(5+3)/(5+1)=8/6
a5=(6+3)/(6+1)=9/7
a6=(7+3)/(7+1)=10/8

ans:5/3,3/2,7/5,4/3,9/7,5/4

But, wait. In the secendon problem, $\displaystyle a_2=\frac{\frac{5}{3}+3}{\frac{5}{3}+1}$ .
$\displaystyle a_3=\frac{a_2+3}{a_2+1}\mbox{ and }a_4=\frac{a_3+3}{a_3+1}\mbox{ and }a_5=\frac{a_4+3}{a_4+1}\mbox{ and etc...}$
{$\displaystyle a_n$}$\displaystyle ^{6}_{n=0}$ if $\displaystyle a_{n+1}$ = ($\displaystyle a_n$+3)/($\displaystyle a_n$+1)