A) \[\frac{n{{(n+1)}^{2}}}{2}\]
B) \[\frac{1}{2}{{n}^{2}}(n+1)\]
C) \[n{{(n+1)}^{2}}\]
D) None of these
Correct Answer: B
Solution :
When \[n\] is odd, the last term. \[i.e.,\] the \[{{n}^{th}}\]term will be \[{{n}^{2}}\] in this case \[n-1\] is even and so the sum of the first \[n-1\] terms of the series is obtained by replacing \[n\] by \[n-1\] in the given formula and so is \[\frac{1}{2}(n-1){{n}^{2}}\]. Hence the sum of the n terms = (the sum of \[n-1\] terms) + the \[{{n}^{th}}\] term \[=\frac{1}{2}(n-1){{n}^{2}}+{{n}^{2}}=\frac{1}{2}(n+1){{n}^{2}}\]. Trick: Check for\[n=1,\ 3\]. Here \[{{S}_{1}}=1,\ {{S}_{3}}=18\] which gives (b).
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