A) \[\frac{n+1}{2}\]
B) \[\frac{n-1}{2}\]
C) \[\frac{{{n}^{2}}+1}{2}\]
D) \[\frac{{{n}^{2}}-1}{2}\]
Correct Answer: A
Solution :
Given series \[\frac{1}{1}+\frac{1+2}{2}+\frac{1+2+3}{3}+......\] So \[{{n}^{th}}\]term of series is given by \[{{T}_{n}}=\frac{1+2+3+.......+n}{n}=\frac{\frac{1}{2}n(n+1)}{n}=\frac{n+1}{2}\]
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