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
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help with fraction sequences 
i dont know wat 2 do so cud u plz explain and show me how 2 do it. thanks muchly 
![\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}](http://www.mathhelpforum.com/math-help/latex2/img/3dbe56201fdf8765bcd651ca1a3450cb-1.gif "\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}")
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