A) \[\frac{m(n+3)}{4}\]
B) \[\frac{(n+3)}{3}\]
C) \[\frac{n(n-3)}{3}\]
D) \[\frac{n(n-3)}{4}\]
Correct Answer: A
Solution :
[a] \[{{n}^{th}}term\,\,or\text{ }{{T}_{n}}=\frac{1+2+3+...+n}{n},\] \[{{S}_{n}}=\sum{{{T}_{n}}=\sum{\frac{1+2+...+n}{n}=\sum{\frac{n(n+1)}{2n}}}}\] \[\Rightarrow \sum{\frac{n+1}{2}=\frac{1}{2}(\sum{n}+\sum{1})=\frac{1}{2}\left[ \frac{n(n+1)}{2}+n \right]}\] \[=\frac{n(n+1)+2n}{4}=\frac{n(n+3)}{4}\]
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