For all n in mathbb{N},if 1^2+2^2+3^2+ldots+n | vedclass

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(A) We know that the sum of the squares of the first $n$ natural numbers is given by the formula: $S_n = 1^2+2^2+3^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{

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$\frac{n^3}{3}$

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(A) We know that the sum of the squares of the first $n$ natural numbers is given by the formula: $S_n = 1^2+2^2+3^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6}$ Expanding this expression,we get: $S_n = \frac{2n^3+3n^2+n}{6} = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$ Since $n \in \mathbb{N}$,$n \ge 1$,therefore $\frac{n^2}{2} + \frac{n}{6} > 0$. Thus,$S_n = \frac{n^3}{3} + (	ext{positive terms}) > \frac{n^3}{3}$. Comparing this with the given inequality $S_n > x$,we find that $x = \frac{n^3}{3}$.

For all $n \in \mathbb{N}$,if $1^2+2^2+3^2+\ldots+n^2 > x$,then $x=$

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