# Thread: help with fraction sequences

1. ## help with fraction sequences

i need help with these problems coz my maths teacher has my book i dont know wat 2 do so cud u plz explain and show me how 2 do it. thanks muchly

*1 write an expression for the nth term of the sequence:
1/3 , 2/5 , 3/7 , 4,9 , 5/11 ..............

*2 the nth term is n/n^2+1

the 1st term in the sequence is 1/2
what are the next 3 terms?

2. help with fraction sequences please xox

The Term n :Tn of an arithmetic sequence can be given by
Tn = a + (n-1)d
where n is the nth term
a is the first term and d is the common difference ( ie any term less the previous term)

so T1 = a + (1-1)d=a
T2 = a +(2-1)d= a+d
T3 = a + (3-1)d = a+2d and so on

In your problem note that the numerator of any Tn is n

The denominator : first term is 3 ; common difference is 2 so denominator is Tn = a + (n-1)d = 3 +2(n-1)

numerator/denominator Tn = n/( 3 + 2(n-1))

*2 Substitution

Tn = n/ (n^2 +1)
then for T1 replace each n with 1 : 1/(1^2 +1 ) = 1/2
similarly for next 3 terms T2, T3 and T4

3. Hello, rachel27!

Are you really having trouble with these?

Write an expression for the $\displaystyle n^{th}$ term of the sequence:

. . $\displaystyle \frac{1}{3}\quad \frac{2}{5}\quad \frac{3}{7}\quad\frac{4}{9}\quad\frac{5}{11}\quad\ hdots$
Can you see the pattern of the numerators and denominators?

. . $\displaystyle \begin{array}{c||c|c|c|c|c} n & 1 & 2 & 3 & 4 & 5 \\ \hline \hline \text{num'r} & 1 & 2 & 3 & 4 & 5 \\ \hline \text{den'r} & 3 & 5 & 7 & 9 & 11 \\ \hline \end{array}$

It should be obvious . . .

. . The $\displaystyle n^{th}$ numerator is $\displaystyle n.$

. . The $\displaystyle n^{th}$ denominator is: $\displaystyle 2n+1$

Therefore, the $\displaystyle n^{th}$ fraction is: .$\displaystyle \frac{n}{2n+1}$

2) The $\displaystyle n^{th}$ term is $\displaystyle \frac{n}{n^2+1}$

The first term is $\displaystyle \frac{1}{2}$

What are the next 3 terms?
How hard can this be?

$\displaystyle f(n) \:=\:\frac{n}{n^2+1}$

. . $\displaystyle \begin{array}{cccccc}f(1) &=&\dfrac{1}{1^2+1} &=& \dfrac{1}{2} \\ \\[-3mm] f(2) &=& \dfrac{2}{2^2+1} &=& \dfrac{2}{5} \\ \\[-3mm] f(3) &=& \dfrac{3}{3^2+1} &=& \dfrac{3}{10} \\ \\[-3mm] f(4) &=& \dfrac{4}{4^2+1} &=& \dfrac{4}{17} \end{array}$