Let P(n): 1^2+2^2+3^2+ldots+n^2 = frac{6(n-1) | vedclass

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(C) We know that the sum of the first $n$ squares is given by $1^2+2^2+3^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6} = \frac{2n^3+3n^2+n}{6}$.Given $P(n): 1

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$P(n)$ is true for all $n \leq 2020$

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(C) We know that the sum of the first $n$ squares is given by $1^2+2^2+3^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6} = \frac{2n^3+3n^2+n}{6}$.Given $P(n): 1^2+2^2+3^2+\ldots+n^2 = \frac{6(n-1)(n-2) \ldots(n-2020) + 2n^3+3n^2+n}{6}$.Substituting the standard sum formula into the given equation,we get:$\frac{n(n+1)(2n+1)}{6} = (n-1)(n-2) \ldots(n-2020) + \frac{n(n+1)(2n+1)}{6}$.This equality holds if and only if $(n-1)(n-2) \ldots(n-2020) = 0$.This product is zero when $n \in \{1, 2, 3, \ldots, 2020\}$.Therefore,$P(n)$ is true for all $n \leq 2020$.

Let $P(n): 1^2+2^2+3^2+\ldots+n^2 = \frac{6(n-1)(n-2) \ldots(n-2020)+2n^3+3n^2+n}{6}$,for all $n \in N$. Then which of the following is correct?

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