A) \[n\,(n-1)\]
B) \[\frac{n\,(n-1)}{2}\]
C) \[n\,(n+1)\]
D) \[\frac{n\,(n+1)}{2}\]
Correct Answer: D
Solution :
We know that natural numbers are \[1,\ 2,\ 3,\ 4,\ ........n\] are in A.P. Now sum\[=\frac{n}{2}[2\times 1+(n-1)1]=\frac{n}{2}(n+1)\].
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