Let \(P(n): 1^2+2^2+3^2+\ldots+n^2\) \(=\frac{6(n-1)(n-2) \ldots(n-2020)+2 n^3+3 n^2+n}{6}\), for all \(n…
Let \(P(n): 1^2+2^2+3^2+\ldots+n^2\) \(=\frac{6(n-1)(n-2) \ldots(n-2020)+2 n^3+3 n^2+n}{6}\), for all \(n \in \mathbf{N}\). Then which of the following is correct?
\(P(n)\) is true for all \(n \in \mathrm{N}\)
\(P(n)\) is true for all \(h>2020\)
\(P(n)\) is true for all \(n \leq 2020\)
\(P(n)\) is not true for any \(n \in N\)
Solution
Given statement
\(\begin{aligned}
& P(n)=1^2+2^2+3^2+\ldots+n^2= \\
& \frac{6(n-1)(n-2) \ldots(n-2020)+2 n^3+3 n^2+n}{6}
\end{aligned}\)
\(\because\) We know that,
\(\begin{aligned}
& 1^2+2^2+3^2+\ldots+n^2=\frac{n(n+1)(2 n+1)}{6} \\
& \therefore \quad \frac{6(n-1)(n-2) \ldots(n-2020)+2 n^3+3 n^2+n}{6} \\
& \quad=(n-1)(n-2) \ldots(n-2020)+\frac{n(n+1)(2 n+1)}{6}
\end{aligned}\)
will be \(\frac{n(n+1)(2 n+1)}{6}\) if \(n=1,2,3, \ldots ., 2020\) only.
Therefore, \(P(n)\) is true for all \(n \leq 2020\).
Hence, option (c) is correct.