\[2+4+7+11+16+......\]to \[n\] terms = [Roorkee 1977] – McqOptions

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1.	\[2+4+7+11+16+......\]to \[n\] terms = [Roorkee 1977]
A.	\[\frac{1}{6}({{n}^{2}}+3n+8)\]
B.	\[\frac{n}{6}({{n}^{2}}+3n+8)\]
C.	\[\frac{1}{6}({{n}^{2}}-3n+8)\]
D.	\[\frac{n}{6}({{n}^{2}}-3n+8)\]
Answer» C. \[\frac{1}{6}({{n}^{2}}-3n+8)\]

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