A) \[\frac{n\,(n+1)\,(2n+1)}{6}\]
B) \[\frac{{{n}^{2}}(n+1)}{4}\]
C) \[\frac{n\,(n-1)\,(2n-1)}{6}\]
D) \[{{n}^{2}}\]
Correct Answer: C
Solution :
Let \[{{T}_{n}}\] be the nth term of the series \[{{T}_{n}}=2\sum{n}-\sum{1}\] Þ \[{{T}_{n}}=\frac{2n(n+1)}{2}-n={{n}^{2}}\] \[\therefore {{S}_{n}}=\sum\limits_{k=1}^{n}{({{k}^{2}})}=\frac{n(n+1)(2n+1)}{6}\] Hence sum of \[(n-1)\] terms\[{{S}_{n-1}}=\frac{(n-1)\,n\,(2n-1)}{6}\].
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