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Given for every $n \in N \;(1^2-a_{1})+(1^2-a_{2})+.......+(n^2-a_{n})=\large\frac{1}{3}n(n^2-1)\;the\;a_{n}$ equals

$(a)\;n\qquad(b)\;n-1\qquad(c)\;n+1\qquad(d)\;2n$

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A)
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Answer : (a) n
Explanation : For n-1 ,
Subtracting from equation given in question ,
$n^2-a_{n}=\large\frac{1}{3}\;n(n-1)(n+1)-\large\frac{1}{3}\;n(n-1)(n-2)$
$n^2-a_{n}=\large\frac{1}{3}n(n-1)3$
$a_{n}=n\;.$
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