# The $n^{th}$ term of series is $\frac{1}{1}$, $\frac{(1+2)}{2}$, $\frac{(1+2+3)}{3}$......is

$(a)\;\frac{n+1}{2}\qquad(b)\;\frac{n-1}{2}\qquad(c)\;\frac{n^2+1}{2}\qquad(d)\;\frac{n^2-1}{2}$

Answer : $\frac{n+1}{2}$
Explanation : The series is of the form $\sum\;\frac{n}{n},$
$\sum\;n=\frac{n(n+1)}{2}$
$N^{th}\;term=\sum\;\frac{n}{n}=\frac{[n(n+1)/2]}{n}$
$=\frac{n+1}{2}.$